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IN MEMORIAM
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ANALYTIC GEOMETRY
1
A SERIES OF MATHEMATICAL TEXTS
EDITED BY
EARLE RAYMOND HEDRICK
THE CALCULUS
By Ellery Williams Davis and William Charles
Brenke.
PLANE AND SOLID ANALYTIC GEOMETRY
By Alexander Ziwet and Louis Allen Hopkins.
PLANE AND SPHERICAL TRIGONOMETRY WITH
COMPLETE TABLES
By Arthur Monroe Kenyon and Louis Ingold,
PLANE AND SPHERICAL TRIGONOMETRY WITH
BRIEF TABLES
By Arthur Monroe Kenyon and Louis Ingold.
THE MACMILLAN TABLES
Prepared under the direction of Earle Raymond Hedrick.
PLANE GEOMETRY
By Walter Burton Eord and Charles Ammerman.
PLANE AND SOLID GEOMETRY
By Walter Burton Ford and Charles Ammerman.
SOLID GEOMETRY
By Walter Burton Ford and Charles Ammerman.
ANALYTIC GEOMETRY
AND
PRINCIPLES OF ALGEBRA
BY
ALEXANDER ZIWET
PROFESSOR OF MATHEMATICS, THE UNIVERSITY OF MICHIGAN
AND
LOUIS ALLEN HOPKINS
INSTRUCTOR IN MATHEMATICS, THE UNIVERSITY OF MICHIGAN
'Nzta gorft
THE MACMILLAN COMPANY
1913
All rights reserved
COPTBIGHT, 1913,
By the MACMILLAN COMPANY.
Set up and electrotyped. Published November, 1913
NortooolJ ^ttw
J. 8. Cashing Co. — Berwick & Smith Co.
Norwood, Masa., U.S.A.
QA'
PREFACE
The present work combines with analytic geometry a num-
ber of topics traditionally treated in college algebra that
depend upon or are closely associated with geometric repre-
sentation. Through this combination it becomes possible to
show the student more directly the meaning and the useful-
ness of these subjects.
The idea of coordinates is so simple that it might (and per-
haps should) be explained at the very beginning of the study
of algebra and geometry. Keal analytic geometry, however,
begins only when the equation in two variables is interpreted
as defining a locus. This idea must be introduced very gradu-
ally, as it is difficult for the beginner to grasp. The familiar
loci, straight line and circle, are therefore treated at great
length.
Simultaneous linear equations present themselves naturally
in connection with the intersection of straight lines and lead
to an early introduction of determinants, whose broad useful-
ness is most apparent in analytic geometry.
The study of the circle calls for a discussion of quadratic
equations which again leads to complex numbers. The geo-
metric representation of complex numbers will present no
great difficulty because the student is now somewhat familiar
with the idea of variables, of coordinates, and even vectors
(in a plane).
The discussion of the conic sections is preceded by the
study, especially the plotting, of curves of the form y = f{x),
vi PREFACE
where f(x) is a polynomial of the second, third, etc. degree.
In connection with this the solution of numerical algebraic
equations can be given a geometric setting.
In the chapters on the conic sections only the most essential
properties of these curves are given in the text; thus, poles
and polars are discussed only in connection with the circle.
Great care has been taken in presenting the fundamental
problem of finding the slope of a curve. It seemed desirable
and quite feasible to introduce the idea of the derivative (of
a polynomial only) in connection with the discussion of alge-
braic equations. The calculus method of finding the slope of
a conic section has therefore been explained, in addition to
the direct geometric method.
The treatment of solid analytic geometry follows more the
usual lines. But, in view of the application to mechanics,
the idea of the vector is given some, prominence; and the
representation of a function of two variables by contour lines
as well as by a surface in space is explained and illustrated
by practical examples.
The exercises have been selected with great care in order
not only to furnish sufficient material for practice in algebraic
work but also to stimulate independent thinking and to point
out the applications of the theory to concrete problems. The
number of exercises is sufficient to allow the instructor to
make a choice.
To reduce the course presented in this book to about one
half its extent, the parts of the text in small type, the chap-
ters on solid analytic geometry, and the more difficult prob-
lems throughout may be omitted.
ALEXANDER ZIWET,
L. A. HOPKINS,
E. R. HEDRICK, Editor.
CONTENTS
PLANE ANALYTIC GEOMETRY
PAGES
Chapter I. Coordinates . 1-22
Chapter II. The Straight Line 23-38
Chapter III. Simultaneous Linear Equations — Determi-
nants ' . . . 39-57
Part I. Equations in Two Unknowns — Determi-
nants of Second Order .... 39-45
Part II. Equations in Three Unknowns — Determi-
nants of Third Order .... 46-57
Chapter IV. Relations between Two or More Lines . . 58-69
Chapter V. Permutations and Combinations — Determi-
nants of any Order 70-86
Chapter VI. The Circle — Quadratic Equations . 87-109
Chapter VII. Complex Numbers 110-130
PartL The Various Kinds of Numbers . . . 110-116
Part II. Geometric Interpretation of Complex Num-
bers . . 117-130
Chapter VIII. Polynomials — Numerical Equations . . 131-168
PartL Quadratic Function — Parabola . . . 131-142
Part IL Cubic Function . . . . . . 143-147
Part III. The General Polynomial .... 148-157
Part IV. Numerical Equations 158-168
Chapter IX. The Parabola 169-197
Chapter X. Ellipse and Hyperbola . . . . 198-222
vii
viii CONTENTS
PAGES
Chapter XI. Conic Sections — Equation of Second Degree 223-247
Part I. Definition and Classification .... 223-231
Part II. Reduction of General Equation . . . 232-247
Chapter XII. Higher Plane Curves 248-276
Part I. Algebraic Curves 248-253
Part II. Special Curves — Defined Geometrically or
Kineniatically 254-260
Part III. Special Transcendental Curves . . . 261-265
Part IV. Empirical Equations 266-276
SOLID ANALYTIC GEOMETRY
Chapter XIII. Coordinates 277-291
Chapter XIV. The Plane and the Straight Line . . 292-316
Part I. The Plane 292-306
Part II. The Straight Line 307-316
Chapter XV. The Sphere 317-331
Chapter XV I. QUadric Surf aces — Other Surf aces . . 332-355
Appendix — Note on Numerical Multiplication and Division 356-367
Answers 359-364
Index 365-369
ANALYTIC GEOMETRY
PLANE ANALYTIC aEOMETRY
CHAPTER I
COORDINATES
1. Location of a Point on a Line. The position of a point
P (Fig. 1) on a line is fully determined by its distance OP
from a fixed point on the line, if we know on which side of
O the point P is situated (to the right or to the left of in
Fig. 1). Let us agree, for instance, to count distances to the
f
Fig. 1
right of as positive, and distances to the left of as negative ;
this is indicated in Fig. 1 by the arrowhead which marks the
positive sense of the line.
The fixed point is called the origin. The distance OP,
taken with the sign + if P lies, let us say, on the right, and
with the sign — when P lies on the opposite side, is called
the abscissa of P.
It is assumed that the unit in which the distances are
measured (inches, feet, miles, etc.) is known. On a geographi-
cal map, or on a plan of a lot or building, this unit is indicated
by the scale. In Fig. 1, the unit of measure is one inch, the
abscissa of P is +2, that of Q is — 1, that of P is — 1/3.
B 1
2 PLANE ANALYTIC GEOMETRY [I, § 2
2. Determination of a Point by its Abscissa. Let us select,
on a given line, an arbitrary origin 0, a unit of measure, and a
definite sense as positive. Then any real number, such as 5,
— 3, 7.35, — V2, regarded as the abscissa of a point F, fully
determines the position of P on the line. Conversely, every
point on the line has one and only one abscissa.
The abscissa of a point is usually denoted by the letter x,
which, in analytic geometry as in algebra, may represent any
real or complex number.
To represent a real point the abscissa must be a real number.
If in any problem the abscissa a; of a point is not a real num-
ber, there exists no real point satisfying the conditions of the
problem.
EXERCISES
1. "What is the abscissa of the origin ?
2. With the inch as unit of length, mark on a line the points whose
abscissas are : 3, —2, VS, — 1.25, — V5, |, — i
3. On a railroad line running east and west, if the station B is 20 miles
east of the station A and the station C is 33 miles east of A, what are the
abscissas of A and C for B as origin, the sense eastward being taken as
positive ?
4. On a Fahrenheit thermometer, what is the positive sense ? What
is the unit of measure ? What is the meaning of the reading 66° ?
What is meant by — 7° ?
5. A water gauge is a vertical post carrying a scale ; the mean water
level is generally taken as origin. If the water stands at -|- 7 on one day
and at —11 the next day, the unit being the inch, how much has the
water fallen ?
6. If xu X2 (read : x one, x two) are the abscissas of any two points
Pi, P2 on a given line, show that the abscissa of the midpoint between
Pi and P2 is ^ (xi + 3^2) • Consider separately the cases when Pi, P2 lie
on the same side of the origin and when they lie on opposite sides.
I, § 3] COORDINATES 3
3. Ratio of Division. A segment AB (Fig. 2) of a straight
line being given, it is shown in elementary geometry how to
find the point C that divides
AB in a given ratio k. Thus,
if it = I, the point G such that
AC^2
AB 5
is found as follows. On any
line through A lay off AD = 2 and AE = 5 ; join B and E.
Then the parallel to BE through D meets AB at the required
point C.
Analytically, the problem of dividing a line in a given ratio
is solved as follows. On the line AB (Fig. 3) we choose a
point as origin and assign a positive sense. Then the
abscissas Xj of A and X2 of B are known. To find a point G
r — 1
^:g:":Z-> '
Fig. 3
which divides AB in the ratio of division k = AG/AB, let us
denote the unknown abscissa of G by x. Then we have
AG=x — Xi, AB = X2 — Xi',
hence the abscissa x oi G must satisfy the condition
H/2 — iCj
whence
yj ^^ yJj ~j~ /t ( «^2 """ *yiy }
or, if we write Ax (read : delta x) for the " difference of the
a^s," I.e. Ax = X2 — Xi,
x = Xi-\-k ' Ax.
Thus, if the abscissas of A and B are 2 and 7, the abscissas
4 PLANE ANALYTIC GEOMETRY [I, § 3
of the points that divide AB in the ratios |, i, |, | are 3, 4^,
8, 9^, respectively. Check these results by geometric con-
struction.
If the segments AC and AB have the same sense, the divi-
sion ratio k is positive. For example, in Fig. 3, the point O
lies between A and B ; hence the division ratio fc is a positive
proper fraction. If the division ratio k is negative, the seg-
ments AC and AB must have opposite sense, so that B and C
lie on the opposite sides of A.
If the abscissas of A and B are again 2 and 7, the abscissa
xof C when A; = 2, - 1, - f, - .2 will be 12, - 3, 0, 1, respec-
tively. Illustrate this by a figure, and check by the geometric
construction.
4. Location of a Point in a Plane. To locate a point in
a plane, that is, to determine its position in a plane, we may
proceed as follows. Draw two lines at right angles in the
plane ; on each of these take the point of intersection O as
origin, and assign a definite positive sense to each line, e.g. by
marking each line with an arrowhead. It is usual to mark
the positive sense of one line by affixing the letter x to it, and
the positive sense of the other line by
affixing the letter ?/ to it, as in Fig. 4.
These two lines are then called the axes
of coordinates, or simply the axes. We
distinguish them by calling the line Ox the
a>axis, or axis of abscissas, and the line Oy
the ?/-axis, or axis of ordinates. Now project the point P on
each axis, i.e. let fall the perpendiculars PQ, PR from P on
the axes. The point Q has the abscissa OQ = x on the axis Ox.
The point R has the abscissa OR = y on the axis Oy. The
distance OQ = RP=x is called the abscissa of P, and
y
B
--,P
r
j
y\
X
1
JC
~~D
Q
Fig.
I
I, §6j
COORDINATES
y
n P'r—
/
1
1
^
1
! X
I ' ^
1
— jp-
m \
JY
p"'
OR = QP = 2/ is called the ordinate of P. The position of the
point P in the plane is fully determined if its abscissa x and
its ordinate y are both given. The two numbers x, y are also
called the coordinates of the point P.
5. Signs of the Coordinates. Quadrants. It is clear
from Fig. 4 that x and y are the perpendicular distances of the
point P from the two axes. It should be observed that each
of these numbers may be positive or
negative, as in § 1.
The two axes divide the plane into
four compartments distinguished as in
trigonometry as the first, second, third,
and fourth quadrants (Fig. 5). It is
readily seen that any point in the first
quadrant has both its coordinates posi-
tive. What are the signs of the coordi-
nates in the other quadrants ? What are the coordinates of the
origin ? What are the coordinates of a point on one of the
axes ? It is customary to name the abscissa first and then
the ordinate ; thus the point (—3, 5) means the point whose
abscissa is — 3 and whose ordinate is 5.
Every point in the plane has two definite real numbers as co-
ordinates; conversely, to every pair of real numbers corresponds
one and ordy one point of the jiilane.
Locate the points: (6, -2), (0, 7), (2-V3, f), (-4, 2V2),
(-5,0).
6. Units. It may sometimes be convenient to choose the
unit of measure for the abscissa of a point different from the
unit of measure for the ordinate. Thus, if the same unit, say
one inch, were taken for abscissa and ordinate, the point (3, 48)
might fall beyond the limits of the paper. To avoid this we
6
PLANE ANALYTIC GEOMETRY
[I, §6
may lay off the ordinate on a scale of i inch. When different
units are used, the unit used on each axis should always be
indicated in the drawing. ^ When nothing is said to the con-
trary, the units for abscissas and ordinates are always under-
stood to be the same.
7. Oblique Axes. The position of a point in a plane can
also be determined with reference to two axes that are 7iot at
right angles ; but the angle <o between these
axes must be given (Fig. 6). The abscissa
and the ordinate of the point P are then / y/
0/a> X /
the segments OQ = x, OB = y cut off on /\ Jg
the axes by the parallels through P to the
axes. If o) = |^7r, i.e. if the axes are at
right angles, we have the case of rectangular coordinates
discussed in §§4, 5. In what follows, the axes are always
taken at right angles unless the contrary is definitely
stated.
8. Distance of a Point from the Origin.
For the distance r = OP (Fig. 7) of the point
P from the origin O we have from the right-
angled triangle OQP:
Fig. 7
p
where x, y are the coordinates of P.
If the axes are oblique (Fig. 8), with the angle
xOy = (a^ we have, from the triangle OQP, in
which the angle at Q is equal to ir — w,* by the
cosine law of trigonometry,
Fig. 8
r = Vx2 -\-y2 — 2 xy cos (tt — w) = Vx^ + y"-^ + 2 xy cos w.
* In advanced mathematics, angles are generally measured in radians, the
symbol tt denoting an angle of 180^.
I, § 9] COORDINATES 7
Notice that these formulas hold not only when the point P
lies in the first quadrant, but quite generally wherever the
point P may be situated. Draw the figures for several cases.
9. Distance between Two Points. By Fig. 9, the distance
d = PiP2 between two points Pi{xi, y^ and ^2(^2? 2/2) can be
found if the coordinates of the two points
are given. For in the triangle P1QP2 we ^
have
PiQ = X2-Xi, QPs = 2/2 - 2/1 ;
hence Fio. 9
(1) e« = V(i»2-a:^i)2 + (2/2-2/1)2.
If we write Ax (§ 3) for the " difference of the aj's " and Ay
for the ^' difference of the ^'s ", i.e.
Ax = X2 — Xi and Ay = 2/2 — 2/1 >
the formula for the distance has the simple form
(2) rZ = V(Ai»)2-|-(Ai/)2;
or, in words,
The distance between any two points is equal to the square root
of the sum of the squares of the differences between their corre-
sponding coordinates.
Draw the figure showing the distance between two points
(like Fig. 9) for various positions of these points and show
that the expression for d holds in all cases.
Show that the distance between two points Pi {xi, ?/i), P2 (0:2, 2/2) when
the axes are oblique, with angle w, is
d = V{x2 - xi)2+ (2/2 - yi)'^ + 2(X2 - xi) (2/2 - y\) cos w
= \/( Ax)2 + (Ay)2 + 2 Ax . Ay . cos w.
8
PLANE ANALYTIC GEOMETRY
[I, § 10
10. Ratio of Division. If two points P^ {x^ , 2/1) «**f^ ^2 fe 2/2)
are given by their coordinates, the coordinates x, y of any point
Pon the line P1P2 can be found if the division ratio P^P/P^P^ = k
is known in lohich the point P divides the segment P^P^. Let Q^ ,
Q2, Q (Fig. 10), be the projections oi P^, P2, P on the axis Ox ;
then the point Q divides Q1Q2 in the same ratio k in which
P divides PiP^- Now as OQi = aji,
0^2= ^2) OQ = X, it follows from § 3
that
X=Xi -\-k(X2— Xi).
In the same way we find by projecting
Pi, P2f P on the axis Oy that
Fig. 10
2/ = 2/i ^-^•0/2-2/l)•
Thus, the coordinates x, y oi P are found expressed in terms
of the coordinates of P^ , Po and the division ratio k. Putting
again X2 — Xi = Ax, 2/2 — 2/1 = ^2/ > we may also write
x = Xi-}-k' Ax, y = yi-\-k'Ay.
Here again the student should convince himself that the
formulas hold generally for any position of the two points, by
selecting numerous examples. He should also prove, from a
figure, that the same expressions for the coordinates of the
point P hold for oblique coordinates.
As in § 3, if the division ratio k is negative, the two
segments P1P2 and P^P must have opposite sense, so that
the points P and Pg must lie on opposite sides of the
point Pi.
Find, e.g., the coordinates of the points that divide the seg-
ment joining (— 4, 3) to (6, — 5) in the division ratios k = ^,
k = 2, fc=— 1, k = — 1, and indicate the four points in a
figure.
I J 11] • COORDINATES 9
11. Midpoint of a Segment. The midpoint P of a segment
P1P2 has for its coordinates the arithmetic means of the corre-
sponding coordinates of P^ and P^ ; that is, if x-^ , 2/1 are the co-
ordinates of Pi, 0-2, 2/2 those of P2, the division ratio being
A; = I", the coordinates of the midpoint P are (§ 10)
a; = a?! -j- "2" (^2 ^1) = 2 (^1 1 •^2)5
2/ = ^1 + i (2/2 - 2/1) = i (2/1 + 2/2).
EXERCISES
1. With reference to the same set of axes, locate the points (6, 4),
(2, - i). (- 6.4, - 3.2), (-4, 0), (- 1, 5), (.001, - 4.01).
2. Locate the points (-3,4), (0,-1), (6, - V2), (1,-10^),
(0,a), (a, 6), (3, -2), (-2, v^).
3. If a and 6 are positive numbers, in what quadrants do the follow-
ing points lie : (a, — 6), (6, a), (a, a), (— &, &), (— &, — a)?
4. Show that the points (a, 6) and (a, — 6) are symmetric with
respect to the axis Ox ; that (a, 6) and (—a, 6) are symmetric with re-
spect to the axis Oy ; that (a, 6) and (— a, — 6) are symmetric with
respect to the origin.
5. In the city of Washington the lettered streets (A street, B street,
etc.) run east and west, the numbered streets (1st street, 2d street, etc.)
north and south, the Capitol being the origin of coordinates. The axes
of coordinates are called aivenues ; thus, e.gr., 1st street north runs one
block north of the Capitol. If the length of a block were 1/10 mile, what
would be the distance from the corner of South C street and East 5th
street to the corner of North Q street and West 14^h street ?
6. Prove that the points (6, 2), (0, - 6), (7, 1) lie on a circle whose
center is (3, — 2).
7. A square of side s has its center at the origin and diagonals coin-
cident with the axes ; what are the coordinates of the vertices ? of the
midpoints of the s.ides ? ...
8. If a point moves jjarallel to the axis Oy, which of its coordinates
remains constant ?
10 PLANE ANALYTIC GEOMETRY [I, § U
9. In what quadrants can a point lie if its abscissa is negative ? its
ordinate positive ?
10. Find the coordinates of the points which trisect the distance be-
tween the points (1, — 2) and (— 3, 4).
11. To what point must the hne segment drawn from (2, —3) to
(—3, 5) be extended so that its length is doubled ? trebled ?
12. The abscissa of a point is — 3, its distance from the origin is 5 ;
what is its ordinate ?
13. A rectangular house is to be built on a corner lot, the front, 30 ft.
wide, cutting off equal segments on the adjoining streets. If the house is
20 ft. deep, find the coordinates (with respect to the adjoining streets) of
the back corners of the house.
14. A baseball diamond is 90 ft. square and pitcher's plate is 60 ft.
from home plate. Using the foul lines as axes, find the coordinates of
the following positions :
(a) pitcher's plate ;
(6) catcher 8 ft. back of home plate and in line with second base ;
(c) base runner playing 12 ft. from first base ;
(d) third baseman playing midway between pitcher's plate and third
base (before a bunt) ;
(e) right fielder playing 90 ft. from first and second base each.
16. How far does the ball ^o in Ex. 14 if thrown by third baseman
in position (d) to second base ?
16. If right fielder (Ex. 14) catches a ball in position (e) and throws
it to third base for a double play, how far does the ball go ?
17. A park 600 ft. long and 400 ft. wide has six lights arranged in a
circle about a central light cluster. All the lights are 200 ft. apart, and
the central cluster and two others are in a line parallel to the length of
the park. What are the coordinates of all the lights with respect to two
boundary hedges ?
18. With respect to adjoining walks, three trees have coordinates
(30 ft., 8 ft.), (20 ft., 45 ft.), (- 27 ft., 14 ft.), respectively. A tree is to
be planted to form the fourth vertex of a parallelogram; where should it
be placed ? (Three possible positions ; best found by division ratio.)
I, § 12]
COORDINATES
11
Fig. 11
12. Area of a Triangle with One Vertex at the Origin.
Let one vertex of a triangle be the origin, and let the other
vertices be P^ {x^, 2/1) and P^ (x^, y^. Draw through P^ and
P2 lines parallel to the axes (Fig. 11). The
area A of the triangle is then obtained by
subtracting from the area of the circum-
scribed rectangle the areas of the three non-
shaded triangles ; i.e.
A = x{y^ - i a;i?/i -\x^^-\ (x^ -x^ {y^ - y^)
= i{^iy2 - a^22/i).
This formula gives the area with the sign -|- or — according
as the sense of the motion around the perimeter OP1P2O is
counterclockwise (opposite to the rotation of the hands of a
clock) or clockwise.
For numerical computation it is most convenient to write
down the coordinates of the two points thus :
^1 2/1
«2 2/2
and to take half the difference of the crosswise products. The
formula is therefore often written in the form
=i
X,
X,
2/2
^1 Vi
X
I 2/2
where the symbol
stands for x^yc^—x-^i, and is called a determinant (of the second
order).
Thus, the area of the triangle formed by the origin with the
pair of points (4, 3) and (2, 5) is
. 4 3
2 5
=:i(4x5-2x3) = 7.
12
PLANE ANALYTIC GEOMETRY
[I, § 13
13. Translation of Axes. Instead of the origin and the
axes Ox, Oy (Fig. 12), let us select a new origin 0' (read : O
prime) and new axes 0'x\ O'y', parallel to the old axes. Then
any point P whose coordinates with reference to the old axes
are OQ=:x, QP = y will have with
reference to the new axes the coordi-
nates 0'Q' = x', Q'P=y'', and the
figure shows that if 7i, k are the co-
ordinates of the new origin, then
X — x' + h,
y = y'-\-k.
yl
y
1
— i — j>
1 y\
k
h \ \ X
Q
Fig. V.
The change from one set of axes to a new set is called a
transformation of coordinates. In the present case, where the
new axes are parallel to the old, this transformation can be
said to consist in a translation of the axes.
14. Area of Any Triangle. Let Pi(aji, y^, P^ix^, 2/2)?
P3 (ajg , 2/3) be the vertices of the triangle (Fig. 13). If we take
one of these vertices, say P3, as new
origin, with the new axes parallel to the ^
old, the new coordinates of Pi , Pg will be :
I
Jb^ Jbo% Jb
Xo X'\
y'i=yi-ys, y'2 = y2-:
Hence, by § 12, the area of the triangle
AAA is
Fig. 13
A = i {x'yy'.,-x'^\) = l [{xi - xs) (2/2 - 2/3) - {xo - x^) (y, - y.,)]
For numerical computation it is best to put down the coordi-
nates of the three points with a 1 after each pair, thus :
I, § 14]
COORDINATES
13
flJl
2/1
1
x^
2/2
1
%
2/.3
1
Then add the three products formed by following the full lines
and subtract the three i^roducts formed by following the dotted
lines as indicated in the accompany-
ing scheme, i.e. form the determinant
(of the third order)
= a^?/2 + a^22/3 + ^32/i - «^32/2 - ^^1 - aJi2/3. /
This is equal to the expression in \
the square brackets above, i.e. to 2 A.
Therefore
^=h
Here as in § 12 the sign of the area is + or — according as
the sense of the motion along the perimeter P^PoP^P^ is coun-
terclockwise or clockwise.
a^
2/1
1
x^
2/2
1
Xz
2/3
1
EXERCISES
1. Find the areas of the triangles having the following vertices :
(a) (1, 3), (5, 2), (4, 6) ; (6) (-2, 1), (2, - 3), (0, - 6) ;
(c) (a, 6), (a, 0), (0, h) ; {d) (4, 3), (6, - 2), (- 1, 5).
2. Show that the area of the triangle whose vertices are (7, — 8),
(— 3, 2), (—5, —4) is four times the area of the triangle formed by-
joining the midpoints of the sides.
3. Find the area of the quadrilateral whose vertices are (2, 3), (— 1,
-1), (-4,2), (-3,6).
4. Find the area of the triangle whose vertices are (a, 0), (0, 6),
C-c, -c).
5. Find the area of the triangle (1, 4), (3, -2), (-3, 16). What
does your result show about these points ?
14
PLANE ANALYTIC GEOMETRY
[I, § 14
6. Find the area of the triangle (a, h + c), (6, c + a), (c, a + h).
What does the result show whatever the values of a^h^c'}
7. Show that the points (3, 7), (7, 3), (8, 8) are the vertices of an
isosceles triangle. What is its area ? Show that the same is true for the
points (a, 6), (6, a), (c, c), whatever a, 6, c, and find the area,
8. Find the perimeter of the triangle whose vertices are (3, 7), (2,
— 1), (5, 3). Is the triangle scalene ? What is its area ?
15. Statistics. Related Quantities. If pairs of values
of two related quantities are given, each of these pairs of
Values is represented by a point in the plane if the value of
one quantity is represented by the abscissa and that of the
other by the ordinate of the point. A curved line joining
these points gives a vivid idea of the way in which the two
quantities change. Statistics and the results of scientific ex-
periments are often represented in this manner.
EXERCISES
1. The population of the United States, as shown by the census reports,
is approximately as given in the following table :
Tear
1790
1800
'10
'20
'30
'40
'50
'60
'TO
'80
'90
1900
'10
Millions
4
5
7
10
13
17
23
31
39
50
63
76
92
Mark the points corresponding to the pairs of numbers (1790, 4),
(1800, 5), etc., on squared pager, representing the time on the horizontal
axis and the population vertically. Connect these points by a curved line.
2. From the figure of Ex. 1, estimate approximately the population
of the United States in 1875 ; in 1905 ; in 1915.
3. From the figure of Ex. 1, estimate approximately when the popula-
tion was 25 millions ; 60 millions ; when it will be 100 milhons.
4. Draw a figure to represent the growth of the population of yom'
own State, from the figures given by the Census Reports.
I, § 15]
COORDINATES
15
[Other data suitable for statistical graphs can be found in large quan-
tity in the Census Keports ; in the Crop Reports of the government ; in
the quotations of the market prices of food and of stocks and bonds ; in
the World Almanac ; and in many other books. ]
5. The temperatures on a certain day varied hour by hour as follows :
A.M.
N.
P.M.
Time . .
Temp. . .
6
50
7
52
8
55
9
60
10
64
11
67
12
70
1
72
2
74
3
75
4
74
5
72
6
69
7
65
8
60
9
57
Draw a figure to represent these pairs of values.
6. In experiments on stretching an iron bar, the tension t (in tons)
and the elongation E (in thousandths of an inch) were found to be as
follows :
t (in tons)
JS (in thousandths of an inch)
6
60
8
81
10
103
Draw a figure to represent these pairs of values.
[Other data can be found in books on Physics and Engineering.]
7. By Hooke's law, the elongation ^ of a stretched rod is supposed
to be connected with the tension t by the formula E = c -t, where c is a
constant. Show that if c = 10, with the units of Ex. 6, the values of E
and t would be nearly the same as those of Ex. 6. Plot the values given
by the formula and compare with the figure of Ex. 6.
8. The distances through which a body will fall from rest in a vacuum
in a time t are given by the formula s = 16 t^, approximately, if t is in
seconds and s is in feet. Show that corresponding values of s and t are
2
64
3
144
4
256
5
400
6
576
Draw a figure to represent these pairs of values.
16 PLANE ANALYTIC GEOMETRY [I, § 16
16. Polar Coordinates. The position of a point P in a
plane (Fig. 14) can also be assigned by its distance OP=r
from a fixed point, or pole, 0, and the angle xOP = (f>, made
by the line OP with a fixed line Ox, the polar axis. The dis-
tance r is called the radius vector, the angle <^ the polar angle
(or also the vectorial angle, azimuth, qmpU- j»
tvAie^ or anomaly), of the point P. The ^^^^
radius vector r and the polar angle <^ are O' ^^^ £>
called the polar coordinates of P. ^'^' ^*
Locate the points: (5, \it), (6, |7r), (2, 140°), (7, 307°),
(V5, tt), (4, 0°).
To obtain for every point in the plane a single definite pair of polar
coordinates it is sufficient to take the radius vector r always positive and
to regard as polar angle the positive angle between and 2 tt (0 ^ < 2 tt)
through which the polar axis (regarded as a half-line or ray issuing from
the pole 0) must be turned about the pole O in the counterclockwise sense
to pass through P. The only exception is the pole for which r = 0,
while the polar angle is indeterminate.
But it is not necessary to confine the radius vector to positive values
and the polar angle to values between and 2 7r. A single definite point
P will correspond to every pair of real values of r and 0, if we agree that
a negative value of the radius vector means that the distance r is to be
laid off in the negative sense on the polar axis, after being turned through
the angle 0, and that a negative value of <j> means that the polar axis
should be turned in the clockwise sense.
The polar angle is then not changed by adding to it any positive or
negative integral multiple of 2 tt ; and a point whose polar coordinates are
r, can also be described as having the coordinates — r, (p ± ir.
Locate the points :
(3, -i^), (a, -Itt), (-5, 75°), (-3, -20°).
17. Transformation from Cartesian to Polar Coordinates,
and vice versa. The coordinates OQ = x, QP=zy, defined in
§ 4, are called cartesian coordinates, to distinguish them
I, § 18] COORDINATES 17
from the polar coordinates. The term is derived from the
Latin form, Cartesius, of the name of Rene Descartes, who
first applied the method of coordinates systematically (1637),
and thus became the founder of analytic geometry.
The relation between the cartesian and polar coordinates of
one and the same point P appears from
Fig. 15. We have evidently :
V
^^V X \ X
Q
Fig. 15
x=:r COS <j>y y— ^^^ + 2/^
2/ = rsin<^, ^^ tan</,=:^.
18. Distance between Two Points in Polar Coordinates.
If two points Pi , Po are given by their polar coordinates, r^ ,
<^i and r^ , <^2 ? the distance d = PjPg between
them is found from the triangle OP1P.2 (Fig. 16),
by the cosine law of trigonometry, if we ob-
serve that the angle at O is equal to ± (<^2— <^i) •
d = Vri^ -1-^2^ — 2 rirs cos (<^2 — <^i)-
Fig. Iti
EXERCISES
1. Find the distances between the points : (2, | tt) and (4, | tt) ;
(a, Itt) and (3 a, ^tt).
2. Find the cartesian coordinates of the points (5, ^tt), (6, — -^tt),
(4a7r), (2, Itt), (7, 7r),(6, -tt), (4,0), (-3,60°), (-5, -90^^).
3. Find the polar coordinates of the points (\/3, 1), (— V3, 1), (1, —1),
i-h -i)» (-«' «)•
4. Find an expression for the area of the triangle whose vertices are
(0, 0), (n, 0i), and (ro , 02).
5. Find the area of the triangle whose vertices are (vi , e6i), (r2 , 02)?
(»'3, 03).
c
18
PLANE ANALYTIC GEOMETRY
[I, § 19
6. Find the radius vector of the point P on the Une joining the points
-Pi {ii'i 1 0i) and P-z (r2, ^2) sucli that the polar angle of P is ^(0i + 02) •
7. If the axes are oblique with angle w, what arc the relations existing
between the cartesian and polar coordinates of a point ?
19. Projection of Vectors. A straight line segment AB
of definite length, direction, and sense (indicated by an arrow-
head, pointing from A to B) is called a vector. The projection
A'B' (Figs. 17, 18) of a vector AB on an axis, i.e. on a line I
on which a definite sense has been selected as positive, is the
product of the length (or absolute value) of the vector AB into
the cosine of the angle between the positive senses of the axis and
the vector :
A'B' = AB cos a.
The positive sense of the axis (drawn through the initial point
of the vector) makes with the vector two angles whose sum is
2 IT = 360°. As their cosines
are the same it makes no differ-
ence which of the two angles is
used.
With these conventions it is
readily seen that the sum of the
projections of the sides of an
open polygon on any axis is equal
to the projection of the closing
side on the same axis, the sides of the open polygon being
taken in the same sense around the perimeter. Thus, in Fig. 19,
Fig. 19
I, § 20] COORDINATES 19
the vectors P1P2) A^s? ••• AA are inclined at the angles
Ui, 02, •" a;i to the axis I; the closing line PiPq makes the
angle «with ?; its projection is P'lP'e] and we have
P1P2 cos «! H- P2P3 cos ao 4- P3P4 cos ccg + P4P5 cos a^ -\- P^Pq cos «5
= P'iP'6 = PiP6COsa.
For, if the abscissas of Pj, Pg , • • • Pe measured along I, from
any origin on /, are Xi, X2, ••• iCg , the projections of the
vectors are iCg — iCi , a^g — ajg , etc., so that our equation becomes
the identity :
•^2 — ^ ~r ^3 — ^2 ~r *^4 — -^s "h -^o — ^4 I -^6 — ^5^^ ^6 — "^i*
20. Components and Resultants of Vectors. In physics,
forces, as well as velocities, accelerations, etc., are represented
by vectors because such magnitudes have not only a numerical
value but also a definite direction and
sense. ? "^^v
According to the j^^^i^'^^^^^^ogram law of Z^^-" /
physics, two forces OPi, OP2, acting on ^ fig. 20 '
the same particle, are together equivalent
to the single force OP (Fig. 20), whose vector is the diagonal
of the parallelogram formed with OPi, OP2 as adjacent
sides. The same law holds for simultaneous velocities and
accelerations, and for simultaneous or consecutive rectilinear
translations. The vector OP is called the resultant of OP^
and OP2 , and the vectors OPi , OP2 are called the components
of OP.
To construct the resultant it suffices to lay off from the ex-
tremity of the vector OPi the vector P^P = OP2 ; the closing
line OP is the resultant. This leads at once to finding the
20
PLANE ANALYTIC GEOMETRY
[I, § 20
resultant OP of any num-
ber of vectors, by adding
the component vectors geo-
metrically, i.e. putting them
together endwise succes-
sively, as in Fig. 21, where
the dotted lines need not
be drawn.
By §19, the projection ^'''- ^^
of the resultant on any axis is equal to the sum of the pro-
jections of all the components on the same axis.
EXERCISES
1. The cartesian coordinates ic, y of any point P are the projections of
its radius vector OP on the axes Ox, Oy. (See § 16.)
2. The projection of any vector AB on the axis Ox, is the difference
of the abscissas of A and B ; similarly for Oy.
3. A force of 10 lb. is inclined to the horizon at 60° ; find its hori-
zontal and vertical components.
4. A ship sails 40 miles N. 60° E., then 24 miles N. 45° E. How far
is the ship then from its starting point ? How far east ? How far north ?
5. A point moves 5 ft. along one side of an equilateral triangle, then
6 ft. parallel to the second, and finally 8 ft. parallel to the third side.
What is the distance from the starting point ?
6. The sum of the projections of the sides of any closed polygon on
any axis is zero.
7. If three forces acting on a particle are parallel and proportional to
the sides of a triangle, the forces are in equilibrium, i.e. their resultant is
zero. Similarly for any closed polygon.
8. Find the resultant of the forces OPi , OP2 , OP3 , OP4, 0P&, if
the coordinates of Pi, P2 , P3, P4, P5 , with O as origin, are (3, 1),
(1, 2), (-1, 3), (-2, -2), (2, -2). (Resolve each force into its
components along the axes.)
I, § 21] COORDINATES 21
9. If any number of vectors (in the same plane) , applied at the ori-
gin, are given by the coordinates x, y of their extremities, the length of
the resultant is =V{IiX)'^ -\-{I>yy^ (where 2x means the sum of the ab-
scissas, Sy the sum of the ordinates) , and its direction makes with Ox an
angle a such that tan a = I,y/I>x.
10. Find the horizontal and vertical components of the velocity of a
ball when moving 200 ft./sec. at an angle of 30° to the horizon.
11. Six forces of 1, 2, 3, 4, 5, 6 lb., making angles of 60° each with
the next, are applied at the same point, in a plane ; find their resultant.
12. A particle at one vertex of a square is acted upon by three forces
represented by the vectors from the particle to the other three vertices ;
find the resultant.
21. Geometric Propositions. In using analytic geometry
to prove general geometric propositions, it is generally conven-
ient to select as origin a prominent point in the geometric
figure, and as axes of coordinates prominent lines of the figure.
But sometimes greater symmetry and elegance is gained by
taking the coordinate system in a general position. (See, e.g.,
Exs. 14, 17, 18, below.)
MISCELLANEOUS EXERCISES
1. A regular hexagon of side 1 has its center at the origin and one
diagonal coincident with the axis Ox ; find the coordinates of the vertices.
2. Show by similar triangles that the points (1, 4), (3, — 2), (— 2,
13) lie on a straight line.
3. If a square, with each side 5 units in length, is placed with one
vertex at the origin and a diagonal coincident with the axis Ox, what are
the coordinates of the vertices ?
4. If a rectangle, with two sides 3 units in length and two sides
3 VS units in length, is placed with one vertex at the origin and a diagonal
along the axis Ox, what are the coordinates of the vertices? There are two
possible positions of the rectangle ; give the answers in both cases.
22 PLANE ANALYTIC GEOMETRY [I, § 21
6. Show that the pomts (0, - 1), (-2, 3), (6, 7), (8, 3) are the
vertices of a parallelogram. Prove that this parallelogram is a rectangle.
6. Show that the points (1, 1), (-1, —1), ( + V3, — >/3) are the
vertices of an equilateral triangle.
7. Show that the points (6, 6), (3/2, - 3), (- 3, 12), (- J^^, 3) are
the vertices of a parallelogram.
8. Find the radius and the coordinates of the center of the circle pass-
ing through the three points (2, 3), (-2, 7), (0, 0).
9. The vertices of a triangle are (0, 6), (4, —3), (—5, 6). Find the
lengths of the medians and the coordinates of the centroid of the triangle,
i.e. of the intersection of the medians.
Prove the following propositions :
10. The diagonals of any rectangle are equal.
11. The distance between the midpoints of two sides of any triangle
is equal to half the third side.
12. The distance between the midpoints of the non-parallel sides of a
trapezoid is equal to half the sum of the parallel sides.
13. In a right triangle, the distance from the vertex of the right angle
to the midpoint of the hypotenuse is equal to half the hypotenuse.
14. The line segments joining the midpoints of the adjacent sides of a
quadrilateral form a parallelogram.
15. If two medians of a triangle are equal, the triangle is isosceles.
16. In any triangle the sum of the squares of any two sides is equal
to twice the square of the median drawn to the midpoint of the third side
plus half the square of the third side.
17. The line segments joining the midpoints of the opposite sides of
any quadrilateral bisect each other.
18. The sum of the squares of the sides of a quadrilateral is equal to
the sum of the squares of the diagonals plus four times the square of the
line segment joining the midpoints of the diagonals.
19. The difference of the squares of any two sides of a triangle is equal
to the difference of the squares of their projections on the third side.
20. The vertices (xi, y{), (x^, 2/2), (arg, Vz) of a triangle being given,
find the centroid (intersection of medians).
Vj>A^
CHAPTER II
THE STRAIGHT LINE
22. Line Parallel to an Axis. When the coordinates x, y
of a point P with reference to given axes Ox, Oy are known,
the position of P in the plane of the axes is determined com-
pletely and uniquely. Suppose now
that only one of the coordinates is
given, say, a? = 3 ; what can be said
about the position of the point P?
It evidently lies somewhere on the
line AB (Fig. 22) that is parallel to
the axis Oy and Jias the distance 3
from Oy. Every point of the line AB
has an abscissa x = 3, and every point
whose abscissa is 3 lies on the line AB.
say that the equation aj = 3
^rt^
A
Fig. 22
3 \4 \s
For this reason we
represents the line AB; we also say that a; = 3 is the equation
of the line AB.
More generally, the equation x=a, where a is any real
number, represents that parallel to the axis Oy whose distance
from Oy is a. Similarly, the equation y = h represents a
parallel to the axis Ox.
EXERCISES ^* \
Draw the lines represented by the equations :
1. x=-2. 4. 5x = 7. 7. 3a; + 1 = 0.
2. ic = 0. 6. y = 0. 8. 10-3y = 0.
3. a: = 12.5, 6. 2y=-7. 9. ?/=±2.
23
i^.
24 PLANE ANALYTIC GEOMETRY [II, § 23
23. Line through the Origin. Let us next consider any
line * through the origin 0, such as the line OP in Fig. 23.
The points of this line have the prop-
erty that the ratio y/x of their coordi-
nates is the same, wherever on this
line the point P be taken. This ratio
is equal to the tangent of the angle a ^
made by the line with the axis Ox, Fig. 23
i.e, to what we. shall call the slope of the line. Let us put
tan a = w ;
then we have, for any point P on this line : y/x = m, i.e. :
(1) y = mx.
Moreover, for any point Q, not on this line, the ratio y/x
must evidently be different from tan a, i.e. from m. The equa-
tion y = mx is therefore said to represent the line through O
whose slope is m; and y = 7nx is called the equation of this line.
We mean by this statement that the relation y = mx is satis-
fied by the coordinates of every point on the line OP, and only
by the coordinates of the points on this line. Notice in partic-
ular that the coordinates of the origin 0, i.e. x = 0, y = 0,
satisfy the equation y = mx.
24. Proportional Quantities. Any two values of x are
proportional to the corresponding values of y it y = mx. For,
if (xi , ?/i) and (iCg , 2/2) ^^^ two pairs of values of x and y that
satisfy (1), we have
yi==mxi, y2 = mx2;
* For the sake of brevity, a straight line will here in general be spoken oi
simply as a line ; a line that is not straight will be called a curve.
II, § 24] THE STRAIGHT LINE 25
hence, dividing,
2/1/2/2 = Va?2.
The constant quantity m is called the factor of proportionality.
Many instances occur in mathematics and in the applied
sciences of two quantities related to each other in this man-
ner. It is often said that one quantity y varies as the other
quantity x.
Thus Hooke's Law states that the elongation E of a, stretched
wire or spring varies as the tension t ; that is, E = Jet, where k
is a constant.
Again, the circumference c of a circle varies as the radius r;
EXERCISES
1. Draw each of the lines :
(a)y = 2x. (c) yz=-j\x. (e) 5x+3?/=0. {g)y = -x.
(b) y=~Sx. {d)5y = Sx. {f)y = x. (h)x-y = Q.
2. Show that the equation ax -^-hy =Q can be reduced to the form
y = 7nx, if & :51b 0, and therefore represents a line through the origin.
3. Find the slope of the lines :
(a) x + y=0. (c) Sx_-^y = 0.
(b) x-y = 0. (d) \/2x + y = 0.
4. Draw a line to represent Hooke's Law E = kt, ii k = 10 (see Ex. 7,
p. 15). Let t be represented as horizontal lengths (as is x in § 23) and
let E be represented by vertical lengths (as is ?/ in § 23).
6. Draw a line to represent the relation c = 2 7rr, where c means the
circumference and r the radius of a circle.
6. The number of yards y in a given length varies as the number of
feet / in the same length ; in particular, f=Sy. Draw a figure to
represent this relation.
7. If 1 in. = 2.54 cm., show that c = 2.54 i, where c is the number of
centimeters and i is the number of inches in the same length. Draw a
figure.
26
PLANE ANALYTIC GEOMETRY
[11, § 25
25. Slope Form. Finally, consider a line that does not pass
through the origin and is not parallel to either of the axes of
coordinates (Fig. 24) ; let it intersect the axes Ox, Oy at A,
B, respectively, and let P(x, y) be any other point on it. The
figure shows that the slope m of y
the line, i.e. the tangent of the
angle a at which the line is in- b
clined to the axis Ox, is ^^^^
RP
m = tan a =
or, since i2P= QP
BR'
-QR=QP-OB
y — b
Fig. 24
-bSiXidBR = OQ=:x:
that is,
(2) y = mx 4- b,
where b = OB is called the intercept made by the line on the
axis Oy, or briefly the y-intercept.
The slope angle a at which the line is inclined to the axis Ox
is always understood as the smallest angle through which the
positive half of the axis Ox must be turned counterclockwise
about the origin to become parallel to the line.
26. Equation of a Line. On the line AB oi Fig. 24 take
any other point P' ; let its coordinates be x', y', and show that
y' = mx' + b.
Take the point P' {x', y') outside the line AB and show that
the equation y = mx + 6 is not satisfied by the coordinates x',
y' of such a point.
For these reasons the equation ys=mx-\-b is said to represent
the line ivhose y-intercept is b and ivhose slope is m ; it is also
called the equation of this line. The ^/-intercept OB = b and
the slope m = tan a together fully determine the line.
II, § 26] THE STRAIGHT LINE 27
Every line of the plane can be represented by an equation of the
form
y = mx + b,
excepting the lines parallel to the axis Oy. When the line be-
comes parallel to the axis Oy, both its slope m and its ly-inter-
cept b become infinite. We have seen in § 22 that the equa-
tion of a line parallel to the axis Oy is of the f brm x — a.
Eeduce the equation ^x—2y=5 to the form y = mx-\-b and
sketch the line.
EXERCISES
1. Sketch the lines whose y-intercept is & = 2 and whose slopes are
m = I, 3, 0, — I ; write down their equations.
2. Sketch the lines whose slope is w = 4/3 and whose ^/-intercepts are
0, 1, 2, 5, — 1, — 2, — 6, — 12.2, and write down their equations.
3. Sketch the lines whose equations are :
(a) y=2x+S. (c) y=x-l. (e) x-y=l. (g) 1x-y + l2=0.
(6) y=_ix+l. (d)x-\-y = l. (/)x-2y + 2=0. (/i) 4x + 3?/ + 5=0.
4. Do the points (1, 5), (-2, -1), (3, 7) lie on the line y = 2x-\-Z ?
5. A cistern that already contained 300 gallons of water is filled at the
rate of 1 00 gallons per hour. Show that the amount A of water in the
cistern n hours after filling begins is J. = 100 w+300. Draw a figure to
represent this relation, plotting the values of A vertically, with 1 vertical
space = 100 gallons.
6. In experiments with a pulley block, the pull p in lbs., required to
lift a load I in lbs., was found to be expressed by the equation p = . 15 Z + 2.
Draw this line. How much pull is required to operate the pulley with no
load (i.e. when 1 = 0)?
7. The readings of a gas meter being tested, T, were found in compari-
son with those of a standard gas meter S, and the two readings satisfied
the equation r = 300 + 1.2 S. Draw a figure. What was the reading
T when the reading S was zero ? What is the meaning of the slope of
the line in the figure ?
28 PLANE ANALYTIC GEOMETRY [II, § 27
27. Parallel and Perpendicular Lines. Two lines
y = m-^x + &i , 2/ = '^^2^* + ^2
are obviously parallel if they have the same slope, i.e. if
(3) mi = m^.
Two lines 2/ = mjcc 4- ft^ , ?/ = mga; + h^ are perpendicular if the
slope of one is equal to minus the reciprocal of the slope of
the other, i.e. if
(4) mima = — 1.
For if m2 = tan aj , mg = tan Wg , the condition that mim^ = — 1
gives tan aa = — 1/tan ctj = — cot a^ , whence ots = «i + i t.
(^C EXERCISES
1. Write down the equation of any line : (a) parallel to y = 3 a: — 2,
(6) perpendicular to y = 3 x — 2.
2. Show that the parallel to y = Sx — 2 through the origin isy = S x.
3. Show that the perpendicular to y =zSx — 2 through the origin is
y=-^x.
4. For what value of b does the line y = Sx + b pass through the
point (4, 1) ? Find the parallel to ?/ = 3 x — 2 through the point (4, 1).
6. Find the parallel to y = 5x + 1 through the point (2, 3).
6. Find the perpendicular to y = 2x — 1 through the point (1, 4).
7. What is the geometrical meaning of 61 = 62 in the equations
y — m-iX + ?>i , y = m^x + &2 ?
8. Two water meters are attached to the same water pipe and the water
is allowed to flow steadily through the pipe. The readings B\ and ^2 of the
two meters are found to be connected with the time t by means of the
equations Bi = 2.6t, i?2 = 2.5« + 150,
where i?i and B2 are measured in cubic feet and t is measured in seconds.
Show that the lines that represent these equations are parallel. What
is the meaning of this fact ?
9. The equations connecting the pull p required to lift a load lo is
found for two pulley blocks to be
pi = .05 w; -t- 2, p2 = .05 w + 1.6
Show that the lines representing these equations are parallel. Explain.
II, §29] THE STRAIGHT LINE 29
10. The equations connecting the pull p required to lift a load w is
found for two pulley blocks to be
Pi = .1510 + 1.5, p<i — .05 w + 1.5.
Show that the lines representing these equations are not parallel, but
that the values of pi and p-i are equal when lo = 0. Explain.
28. Linear Function. The equation y = mx+b, when m
and b are given, assigns to every value of x one and only one
definite value of y. This is often expressed by saying that
mx + 6 is a function of x ; and as the expression mx + 6 is of
the first degree in x, it is called Siftinctiori of the first degree or,
owing to its geometrical meaning, a linear function of x.
Examples of functions of x that are not linear are 3 ic^ — 5,
ax^ -\-hx-\-c, x{x — l), 1/x, sin a;, 10"^, etc. The equations
y = 3a^ — 5, y = ax^ -{- bx -\- Cj etc., represent, as we shall see
later, not straight lines but curves.
The linear function y = mx + b, being the most simple kind
of function, occurs very often in the applications. Notice that
the constant b is the value of the function for x = 0. The con-
stant m is the rate of change of y with respect to x.
29. Illustrations. Example 1. A man, on a certain date,
has $10 in bank; he deposits $3 at the end of every week;
how much has he in bank x weeks after date ?
Denoting by y the number of dollars in bank, we have
y = 3x-\-10.
His deposit at any time a; is a linear function of x. Notice
that the coefficient of x gives the rate of increase of this de-
posit ; in the graph this is the slope of the line.
Example 2. Water freezes at 0° C. and 32° F. ; it boils at
100'* C. and at 212° F. ; assuming that mercury expands uni-
formly, i.e. proportionally to the temperature, and denoting
30 PLANE ANALYTIC GEOMETRY [II, § 29
by X any temperature in Centigrade degrees, by y the same
temperature in Fahrenheit degrees, we have
y-S2 212-32 9 . o . oo
If the line represented by this equation be drawn accurately,
on a sufficiently large scale, it could be used to convert centi-
grade temperature into Fahrenheit temperature, and vice versa.
Example 3. A rubber band, 1 ft. long, is found to stretch
1 in. by a suspended mass of 1 lb. Let the suspended mass
be increased by 1 oz., 2 oz., etc., and let the corresponding
lengths of the band be measured. Plotting the masses as ab-
scissas and the lengths of the band as ordinates, it will be
found that the points (x, y) lie very nearly on a straight line
whose equation is y = ^^x -\-l. The experimental fact that
the points lie on a straight line, i.e. that the function is linear,
means that the extension, y — 1, is proportional to the tension j
i.e. to the weight of the suspended mass x (Hooke's Law).
Notice that only the part of the line in the first quadrant,
and indeed only a portion of this, has a physical meaning.
Can this range be extended by using a spiral steel spring ?
Example 4. When a point P moves along a line so as to
describe always equal spaces in equal times, its motion is called
uniform. The spaces x^assed over are then proportional to the
times in which they are described, and the coefficient of pro-
portionality, i.e. the ratio of the distance to the time, is called
the velocity v of the uniform motion. If at the time t = the
moving point is at the distance Sq, and at the time t at the dis-
tance s, from the origin, then
S = SQ-\-Vt.
Thus, in uniform motion, the distance s is a linear function of
the time t, and the coefficient of t is the speed : v = (s — SQ)/t.
II, § 29] THE STRAIGHT LINE 31
Example 5. When a body falls from rest (in a vacuum) its
velocity v is proportional to the time t of falling : v= gt, where
g is about 32 if the velocity is expressed in ft./sec, or 980
if the velocity is expressed in cm./sec.
If, at the time t = 0, the body is thrown downward with an
initial velocity Vq, its velocity at any subsequent time t is
v = Vq + gt.
Thus the velocity is a linear function of t, and the coefficient g
of t denotes the rate at which the velocity changes with the
time, i.e. the acceleration of the falling body.
EXERCISES
1. Draw the line represented by the equation y = f x + 32 of Ex-
ample 2, § 29. What is its slope ? What is the y-intercept ? What is
the meaning of each of these quantities if y and x represent the tempera-
tures in Fahrenheit and in Centigrade measure, respectively ?
2. Kepresent the equation ?/ = j^^ a: + 1 of Example 3, § 29, by a figure.
What is the meaning of the ?/-intercept ?
3. Draw the line s = sq -\- vt of Example 4, § 29, for the values Sq = 10,
^ = 3. What is the meaning ofv? Show that the speed v may be thought
of as the rate of increase of s per second.
4. If, in the preceding exercise, v be given a value greater than 3,
how does the new line compare with the one just drawn ?
6. If, in Ex. 3, v is given the value 3, and so several different values,
show that the lines represented by the equation are parallel. Explain.
6. In experiments on the temperatures at various depths in a mine,
the temperature (Centigrade) T was found to be connected with the
depth d by the equation r= 60 + .01 d, where d is measured in feet.
Draw a figure to represent this equation. Show that the rate of increase
of the temperature was 1° per hundred feet.
7. In experiments on a pulley block, the pull p (in lb.) required to
lift a weight w (in lb.) was found to he p = .03 w ■{■ 0.5. Show that the
rate of increase of p is 3 lb. per hundred weight increase in w.
32 PLANE ANALYTIC GEOMETRY [II, § 30
30. General Linear Equation. The equation
in which A, B, C are any real numbers, is called the general
equation of the first degree in x and y. The coefficients Ay B, C
are called the constants of the equation ; x, y are called the
variables. It is assumed that A and B are not both zero.
The terms Ax and By are of the first degree ; the term C is
said to be of degree zero because it might be written in the
form Cx^ ; this term C is also called the constant term.
Every equation of the first degree,
(5) Ax+By-\-C = 0,
in which A and B are not both zero, represents a straight line;
and conversely, every straight line can be represented by such an
equation. For this reason, every equation of the first degree
is called a linear equation.
The first part of this fundamental proposition follows from
the fact that, when B is not equal to zero, the equation can be
reduced to the form y = mx-^ bhy dividing both sides by B ;
and we know that y = mx -\- b represents a line (§ 25). When
B is equal to zero, the equation reduces to the form x = a,
which also represents a line (§ 22).
The second part of the theorem follows from the fact that
the equations which we have found in the preceding articles
'for any line are all particular cases of the equation
Ax-\-By -\- C = 0.
This equation still expresses the same relation between x
and y when multiplied by any constant factor, not zero. Thus,
any one of the constants A, B, C, if not zero, can be reduced
to 1 by dividing both sides of the equation by this constant.
The equation is therefore said to contain only tivo (not three)
essential constants.
II, § 32]
THE STRAIGHT LINE
33
31. Conditions for Parallelism and for Perpendicularity.
It is easy to recognize whether two lines whose equations are
Ax + By-i-C=0 and A'x -\- B'y + C" = are parallel or per-
pendicular. The lines are parallel if they have the same slope,
and they are perpendicular (§ 27) if the product of their slopes
is equal to —1. The slopes of our lines are — A/B and
— A'/B' ; hence these lines are parallel if — A/B = — A'/B'y
*-e-if A:B = A':B':
and they are perpendicular if
^.^; = -l, Le.ii
B B' '
AA' + BB' = 0.
32. Intercept Form. If the constant term (7 in a linear
equation is zero, the equation represents a line through the
origin. For, the coordinates (0, 0) of the origin satisfy the
equation Ax + By = 0.
If the constant terra C is not equal to zero, the equation
Ax + By -\- C = can be divided by (7 ; it then reduces to the
form
^x + |, + l=0.
If A and B are both different from zero, this can be written :
+
y
- C/A ' - C/B '
or putting — C/A = a, — C/B
(6)
6:
a o
Fig. 25
The conditions A^(), B^^O mean
evidently that the line is not parallel to either of the axes.
Therefore the equation of any line, not passing through the
origin, and not parallel to either axis, can be written in the
34 PLANE ANALYTIC GEOMETRY [II, § 32
form (6). With 2/ = this equation gives a; = a; with x =
it gives y = b. Thus
^ B' A
are the intercepts (Fig. 25) made by the line on the axes Oxy
Oyj respectively (see § 25).
EXERCISES
1. Write down the equations of the line whose intercepts on the
axes Ox^ Oy are 5 and — 3, respectively ; the line whose intercepts are
— I and 7 ; the line whose intercepts are — 1 and — |. Sketch each of
the lines and reduce each of the equations to the form Ax-\-By-{-C=0, so
that A, B, C are integers.
2. Find the intercepts of the lines : Sx — 2y = 1, x + ly-{-l = 0,
— Sx + ^y — 5 = 0. Try to read off the values of the intercepts directly
from these equations as they stand.
■ 3. In Ex. 2, find the slopes of the lines.
4. Prove (6), § 32 by equality of areas, after clearing of fractions.
5. What is the equation of the axis Oy ? of the axis Ox ?
6. What is the value of B such that the line represented by the equa-
tion ix-{- By — li = passes through the point (— 5, 17) ?
7. What is the value of A such that the line Ax -\- 7 y = 10 has its
OS-intercept equal to — 8 ?
8. Reduce each of the following equations to the intercept form (6),
and draw the lines :
(a) Sx-6y-16 = 0. (b) x -{- ^y + 1= 0.
(c) i ^-3y-6 ^^ (d) 5.x = 3x + ?/-10.
x + y
9. Reduce the equations of Ex. 8 to the slope form (2), § 25.
10. Find the equation of the hue of slope passing through the point
(6,-5).
n, §32] THE STRAIGHT LINE 35
XI. What relation exists between the coefficients of the equation
Ax-i- By + C = 0, ii the line is parallel to the line ix — 6y = 8? parallel
to the axis Oy ?
12. Show that the points (- 1, -7), Q, -3), (2,2), (-2, -10)
lie on the same line.
13. Find the area of the triangle formed by the lines x+y=0, x—y=0,
X— a = 0.
14. Show that the line 4(ic — a) + 5(y — &) = is perpendicular to the
line 5 ic— 4 y—10=0 and passes through the point (a, b).
15. A line has equal positive intercepts and passes through (—5, 14).
What is its equation ? its slope ?
16. If a line through the point (6, 7) has the slope 4, what is its
y-intercept ? its a;-intercept ?
17. The Reaumur thermometer is graduated so that water freezes at
0° and boils at 80". Draw the line that represents the reading B of the
Reaumur thermometer as a function of the corresponding reading G of
the Centigrade thermometer.
18. What function of the altitude is the area of a triangle of given
base ?
19. A printer asks 75 f to set the type for a program and 2 ^ per copy
for printing. The total cost is what function of the number of copies
printed ? Draw the line representing the function.
Another printer asks 3 ^ per copy, with no charges for setting the type.
For how many copies would both charge the same ?
20. The sum of two complementary angles a and j3 is ^ tt ; draw the
line representing /3 as a function of a. When a = | tt, what is /3 ?
21. Express the value of a note of § 1000 at the end of the first year as
a function of the rate of interest. At 6% simple interest its value is what
function of the time in years ?
22. Two weights are attached to the opposite ends of a rope that runs
through a double pulley block of which one block is fastened at a height
above ground. If x and y denote the distances of the two weights above the
ground, determine a linear relation between them if a; = 40 when y =
and y = 10 when x = 0.
36 PLANE ANALYTIC GEOMETRY [II, § 33
33. Line through One Point. To find the line of given
slope 7ni through a given point Pi(i«i, 2/1)? observe that the
equation must be of the form (2), viz.
y = TiiiX + b,
since this line has the slope m^. If this line is to pass through
the given point, the coordinates x^, y^ must satisfy this equa-
tion, i.e. we must have
2/1 = ra^^i + &•
This equation determines h, and the value of h so found might
be substituted in the preceding equation. But we can eliminate
h more readily between the two equations by subtracting the
latter from the former. This gives
y-yi = m,{x-x,)
as the equation of the line of slope Wj through Pi{xi, y^.
The problem of finding a line through a given point parallel,
or perpendicular, to a given line is merely a particular case of
the problem just solved, since the slope of the required line can
be found from the equation of the given line (§ 27). If the
slope of the given line is m^ = tan a-^, the slope of any parallel
line is also mj, and the slope of any line perpendicular to it is
mg = tan (ai-\-lir) = — cot a^ = — — .
mi
34. Line through Two Points. To find the line through two
given points, Pi{xi, 2/0? -^2(^2? 2/2)? observe (Fig. 26) that the
slope of the required line is evi- ^
dently
if, as in § 9, we denote by A a;, A?/
the projections of P1P2 011 Ox, Oy\
II, § 34] THE STRAIGHT LINE 37
and as the line is to pass through (xi, y^), we find its equation
by § 33 as V V ^ ^. V , ^ j^^ ,; ^p^
2/2-2/1, .
— W, = (x — cc, ),
a; 2/
1
^'1 2/1
1
^2 2/2
1
y-yi = -~(^-^i)'
The equation of the line through two given points (aj^, yi),
fe 2/2) can also be written in the determinant form
which (§ 14) means that the point (x, y) is such as to form
with the given points a triangle of zero area. By expanding
the determinant it can be shown that this equation agrees with
the preceding equation. A more direct proof will be given
later (§ 49).
EXERCISES
1. Find the equation of the line through the point (—7, 2) parallel
to the line y = Sx.
2. Show that the points (4, —3), (—5, 2), (5, 20) are the vertices of
a right triangle.
3. Find the equation of the line through the point (— 6, — 3) which
makes an angle of 30° with the axis Ox ; 30° with the axis Oy.
4. Does the line of slope | through the point (4, 3) pass through the
.point (—5, -4) ?
5. Find the equation of the line through the point (—2, 1) parallel to
the line through the points (4, 2) and (- 3, — 2).
6. Find the equations of the lines through the origin which trisect
that portion of the line 5 x - 6 y = 60 which lies in the fourth quadrant.
7. What are the intercepts of the line through the points (2, —3),
(-5, 4) ?
38 PLANE ANALYTIC GEOMETRY [II, § 34
8. Show that the equation of the line through the point (a, 6) per-
pendicular to the line Ax-\- By -{- C = is (x — a)/ A = (y — h)/B.
9, Find the equations of the diagonals of the rectangle formed by the
lines ic + a = 0, a; — &=0, ?/ + c = 0, y — cZ = 0.
10. Find the equation of the perpendicular bisector of the line joining
the points (4, —5) and (— 3, 2). Show that any point on it is equally
distant from each of the two given points.
11. Find the equation of the line perpendicular to the line 4x— 3?/+6=0
that passes through the midpoint of (—4, 7) and (2, 2).
12. What are the coordinates of a point equidistant from the points
(2, —3) and ( — 5, 0) and such that the line joining the point to the origin
has a slope 1 ?
13. If the axes are oblique with angle t»7, show that the slope of the
line joining the points P\(x\, y\) and P^ixo, y^) is
(y2 — yi) sin a>
(a;2-a;i) + {y2-yi)cosw
^ 14. If the axes are oblique with angle w, show that the equation of the
line through the point Pi{xi, yi) which makes with the axis Ox the
angle (p, is
sm (w — 0)
Is the coefficient of (x — Xi) the slope of this line ?
15. In an experiment with a pulley-block it is assumed that the rela-
tion between the load I and the pull p required to lift it is linear. Find
the relation Up = 8 when I = 100, and p = 12 when I = 200.
16. In an experiment in stretching a brass wire it is assumed that the
elongation E is connected with the tension t by means of a linear relation.
Find this relation if t = IS lb. when E = .1 in., and i = 58 lb. when
^ = .3 in.
17. A cistern is being filled by water flowing into it at the rate of 30
gallons per second. Assuming that the amount A of water in the cistern
is connected with the time « by a hnear relation, find this relation if
A = 1000 when ( = 10. Hence find A when t = 0.
CHAPTER III
SIMULTANEOUS LINEAR EQUATIONS
DETERMINANTS
PART I. EQUATIONS IN TWO UNKNOWNS
DETERMINANTS OF SECOND ORDER
35. Intersection of Two Lines. The point of intersection
of any two lines is found by solving the equations of the lines as
simultaneous equations. For the coordinates of the point of
intersection must satisfy each of the two equations, since this
point lies on each of the two lines ; and it is the only point
having this property. Find the points of intersection of the
following pairs of lines :
^^^ l3a; + 52/-34 = 0. ^^ \lx + 2y=^0. .
^^^ l5a;-22/ + ll = 0.
The solution of simultaneous linear equations is much
facilitated by the use of determinants. As, moreover, deter-
minants are used to advantage in many other problems (see,
e.g., §§ 12, 14) it is desirable to study determinants systemati-
cally before proceeding with the study of the straight line.
36. Solution of Two Linear Equations. To solve two
linear equations (§ 30),
[ ttga; + &^ = ^2 J
we may eliminate y to find x, and eliminate x to find y. The
elimination of y is done systematically by multiplying the first
39
40
PLANE ANALYTIC GEOMETRY [III, § 36
equation by 63? the second by 61, and then subtracting the
second from the first ; this gives
Likewise, to eliminate x, multiply the first equation by a2, the
second by aj , and subtract the first from the second :
If aib2 — a2&i =^ ^} we can divide by this quantity and thus
find
/f)\ ^ "'1"2 — ^2^1 „. ^V^2 — ^2%
(Z) X = — ; — , y = ; , •
^ a^2 — «20i otiOa — «20i
Observe that the values of x and y are quotients with the
same denominator, and that the numerator of x is obtained
from this denominator by simply replacing a by A;, while the
numerator of y is obtained from the same denominator by
replacing h by k.
This peculiar form of the numerators and denominators of
X and y is brought out more clearly if we agree to write the
common denominator ajftg — ^2^1 in the form of a determinant :
an 60
(3)
as in § 12.
Thus
2
7
-1
4
=2x5-7x3
11;
= -lx2-4x7=-30.
With this notation, the values (2) of x and y are
(4)
a; =
^1 &i
"'2 ^2
%
fc,
aj
fc,
Oi
61
aj
6.
I"
Ill, § 37] SIMULTANEOUS LINEAR EQUATIONS
41
37. General Rule. If a, h, c, d are any four numbers, the
expression a i, i
c ay
which stands for ad — he, is called a determinant, more pre-
cisely, a determinant of the second order because two numbers
occur in each (horizontal) row, as well as in each (vertical)
column. (See § 12.)
The determinant (3) is called the determinant of the equa-
tions (1), § 36.
We can then state the following rule for solving the two
linear equations (1) : If the determinant of the equations is not
equal to zero, x as well as y is the quotient of two determinants ;
the denominator is the sayiie, viz. the determinant of the equa-
tio7is (1) ; the numerator of x is obtained from this denominator
by replacing the coefficients of x by the constant terms, the numer-
ator of y is found from the same denominator by replacing the
coefficients ofy by the constant terms.*
EXERCISES
1. Find the values of the following determinants :
(«)
10 2
3 7
(c?)
12 5
(0
2-1
1 2
h -12
f -§
2. Solve the following equations ; in writing down the solution, begin
with the denominators :
(a-) P^-2y = l,
|2x + 3y + 4 = 0,
^ ^ \Bx-5y-16 = 0.
(ft)
i2x + 7y = 3,
\5x-y = -ll.
l6x-3y-2 = 0,
[y =:4x-l.
* One great advantage of this rule is that the same rule applies to the solu-
tion of any (finite) number of linear equations with the same number of
variables. (See § 74.)
42 PLANE ANALYTIC GEOMETRY [III, § 38
38. Exceptions. The process of § 37 cannot be applied
when the determinant of the equations (1) vanishes, i.e. when
= 0,
that is, when ajj2 = ^2^1 •
For the sake of simplicity we here assume that none of the
four numbers a^, b^, a^, 6, is zero. If any one of them were
zero, we might solve the equation in which it occurs to obtain
the value of one of the variables. With this assumption, the
condition may be written in the form
02 _ 62
ai by
or, denoting the common value of these quotients by m :
a^ = ma^f b^ = mb^,
so that the equations (1) become
a^x + biy = A^i,
maiX + mbiy = k^.
We must now distinguish two cases, according as k^ = m\ or
^2 ^ mfci. In the former case, i.e. if
k.^ = mkx,
the second equation reduces, upon division by m, to the first
equation. Thus, the two equations represent one and the
same relation between x and y, and are therefore not sufficient
to determine x and y separately. We can assign to either
variable an arbitrary value and then find a corresponding
value of the other variable. The equations (1) can then be
said to have an infinite number of solutions.
In the other case, i.e. if
the equations are evidently inconsistent, and there exist no
finite values of x and y satisfying both equations. Thus the
equations \x — 2y = 2, 2 « — 12 v/ = 15 are inconsistent.
Ill, § 40] SIMULTANEOUS LINEAR EQUATIONS
43
39. Geometric Interpretation. All these results about
linear equations can be interpreted geometrically. We have
seen (§ 30) that every linear equation represents a straight
line, and (§ 35) that by solving two such equations we find
the coordinates of the point of intersection of the two lines.
Now two lines in a plane may either intersect, or coincide, or
be parallel. In the first case, they have a single point in com-
mon ; in the second, they have an infinite number of points in
common ; in the third, they have no point in common. The
first case is that of §§ 36, 37 ; the last two cases are discussed in
§ 38. Including the case of coincident lines with that of paral-
lels, we may say that the relation
is the necessary and sufficient condition of parallelism of the
two lines a^x -{- hy = h, a^x + b.jy = k,.
=
40. Elimination. If in the linear equations (1) of § 36 the
constant terms k^, k^ are both zero so that they are
a^x + h^y = 0,
a.2X-\-b2y = 0,
the equations are called homogeneous. Obviously, two homo-
geneous linear equations are always satisfied by the values
x= 0, ^ = 0.
If the determinant of ^ the equations does not vanish, i.e. if
this solution is also found from § 36, and it is the only solution.
But if ai &i
it is found as in § 38 that the equations have an infinite num-
ber of solutions. Conversely, if two homogeneous linear eqiia-
=0,
44
PLANE ANALYTIC GEOMETRY [III, § 40
tions are satisfied by values of x and y that are not both zero, the
determinant of the equations must vanish. For, multiplying the
first equation by 62; and the second by 5^, and subtracting, we
Eliminating a? in a similar manner, we find
(ai&2-a2&i)2/=0.
These equations show that unless x and y are both zero we
must have
% 5i
a^bz — a^b^ = 0, i.e.
a.2 &2
= 0.
This relation is also the result of eliminating x and y between
the two equations. For, if, e.g., jc =/= we may divide both
equations by x and then eliminate y/x between the equations
ai4-6i^ = 0, a2 + &2-=0,
X X
by multiplying the former by 62) the latter by b^, and subtract-
ing. The result is again afiz — ajb^^ = 0. Thus the result of
eliminating the variables between two homogeiieous linear equa-
tions is the determinant of the equations equated to zero. We
shall see later (§ 75) that all the results of the present article
are true for any number of homogeneous linear equations.
Geometrically, two homogeneous linear equations of course
represent two lines through the origin. The vanishing of the
determinant means that the lines coincide so that they have an
infinite number of points in common.
EXERCISES
1. Evaluate the determinants :
(a)
2 5
3 4
>
(&)
7 -3
4 1
5
(c)
1 a
-a 1
'
id)
sin^
cos j9
— cos /S
sin^
; (e)
1
cos/3
30Si8
1
; (/)
ai + a2
at
02
a2 + a3
Ill, §40] SIMULTANEOUS LINEAR EQUATIONS
45
2. Express x^ + y- in the form of a determinant of the second order.
3. Verify that
a2 + 52 aa' + bb'
a
b
2
and that
aa'-\-bb' a'^-hb'^
a' b'
>
a2 + 62 _|_ c2 aa' + bh' + cc'
aa' + bb' + cc' a'^ + 6'2 _ c'2
=
b c
b< c'
2
+
c a
c' a'
2 a
^ a'
b
b'
4. Verify that
\aa' -\-hh' ac' +bd'
\a b\ \a' b'
lea
' + db' cc' +dd'
~
c
d
n
c' d'
(a)
(d)
(a)
(d)
(a)
:3x- 6^-8=0,
[ x-2y + l = 0.
■4x-2y-7 = 0,
2x-Sy + 5 = 0.
|5x-7y + 6 = 0,
l5x-7y+s=0.
(&)
(&)
5. Find the coordinates of the points of intersection of the following
lines ; and check by a sketch :
|6x-7y+ll = 0, j4x+2y-7zz:0, • i2x-by = 3,
[3x+2y-12=0. ^ ^ [3x-8y+4=0. ^^^ | x + Sy=-l.
j4x + 2i/ = 9, I 3x+2i/=0, |2.4x+3.l2/=4.5,
|2x-5y=0. ^^^ |6x-4i/+4=0. ^*'^ [ .8x + 2y=6.2.
6. Do the following pairs of lines intersect, or are they parallel or
coincident ?
I 3x + ?/-6 = 0, I 3x-5?/=0,
I »; + i?/-2=0. ^^^ |]0?/-6x=0.
f2x-62/-4=0 I x + iy = 0,
I x-3y-2=0.'^^\2x + S y = 0.
For what values of s do the following pairs of lines become parallel ?
f4x + sy-15 = 0, j 3sx-8?/-13=0, |7x - Uy + 8 =0,
|2x-7i/+10 = 0. ^ ^ {2x-2s^+15=0. ^^-^ jsx- 2y + s=0.
8. For what values of s do the following pairs of lines coincide ?
3x + 2?/ + 3 = 0, |3x + 6y-5 = 0,
sx —2y -\- s =0. ^^M x + sy — f = 0.
Solve the following equations by determinants :
(«)
(d)
{2 U-3V-.
I?
jX y
25,
5.
(&)
? = 2,
Ix y
ie)
I X2+ ?/2
[2x2- 3y2
x2 2/2 3 '
i+2i=:
25,
5.
(/)
f s = 16 «2 + 100,
^^^ {5s + <2 = 824
1 3
x+y x—y
2 , 5
x-\-y x — y
= - 8,
= 17.
46
PLANE ANALYTIC GEOMETRY [III, § 41
PART XL EQUATIONS IN THREE UNKNOWNS
DETERMINANTS OF THIRD ORDER
41. Solution of Three Linear Equations.
linear equations with three variables x, y, Zy
To solve three
(1)
Kin
a^x + 632/ + Cg^ = k^,
in a systematic way, we might first eliminate z between the
second and third equations (by multiplying the second by C3,
the_ third by Co, and subtracting) ; and then eliminate z between
the third and first equations. We should then have two linear
equations in x and y, which can be solved as in § 36. This
method is long and tedious. But we can find x directly by
multiplying the three given equations respectively by
^2^3 ^3^2 —
63C1 — biC^ =
^
61C2 — b^Ci
and adding the resulting equations. For it is readily verified
that, in the final equation, the coefficients of y and z, viz.
&i
+ &2
^3 C3
63 C
are both zero
&2 ^2
61 Ci
+ 63
Ci
+ C2
+C3
a
Cz
4-a2
\02 C2IJ
fcl
+ A:2
-fA:3
^
62 C2 I ' ^ I &:
We find therefore
^3 C3
&2 C2
^3 C3
i.e. if the coefficient of ic is =/= 0,
^162^3 — fei&3C2 4- ^2^3^! — kzbiC^ 4- k^biCi — A;362Ci
^1^2<^3 ~ ^1^3C2 + 0^2^361 — «2&iC3 + (^5^162 — O362C1
Observe that the numerator is obtained from the denomina-
tor by simply replacing every a by the corresponding k.
-[
:11
x =
Ill, § 42] SIMULTANEOUS LINEAR EQUATIONS
47
It can be shown similarly that y is a, quotient with the same
denominator, and with the numerator obtained from the de-
nominator by replacing every b by the corresponding k ; and that
z is a quotient with the same denominator and the numerator
obtained by replacing every c by the corresponding k.
42. Determinants. The common denominator of x, y, z is
usually written in the form
«! bi Ci
(Xo ^2 2
*3 C3
(2)
and is then called a determinant of the third order. The nine
numbers ai, &i, q, ag, 62? ^2? ^3? ^3? Cg are called its elements ; the
horizontal lines are called the rows, the vertical lines the
columns. The diagonal through the first element a^ is called
the principal diagonal ; that through a^ the secondary diagonal.
By § 41 we have
% b.
Ci
b2
C2
h
C3
h
Ci
C2
= %
h
+ «2
h
+ ^3
h
C3
Ci
C2
Gz
Thus, a determinant of the third order represents a sum of six
terms, each term being a product of
three elements and containing one
and only one element from each row
and from each column.
The most convenient method for
expanding a determinant of the
third order, i.e. for finding the six
terms of which it is the sum, is
indicated by the adjoining scheme.
48
PLANE ANALYTIC GEOMETRY [III, § 42
Draw the principal diagonal and the parallels to it, as in the
figure ; this gives the terms with sign + ; then draw the
secondary diagonal and the parallels to it ; this gives the terms
with sign — . (Compare § 14.)
43. General Rule. When three linear equations, like (1),
§ 41, are given, the determinant (2), § 42, of the coefficients of
X, y, z is called the determinant of the equations. We can now
state the rule for solving the equations (1) when their determi-
nant is different from zero, by the following formulas (compare
§36):
a; =
i.e. each of the variables is the quotient of two determinants; the
denominator in each case is the determinant of the equations, while
the numerator is obtained from this common denominator by re-
placing the coefficients of the variable by the constant terms.
It will be shown in solid analytic geometry that any linear
equation in x, y, z represents a plane. Hence by solving the
three simultaneous equations of § 41 we find the point (or
points) common to three planes.
h
&1
Cl
(Xl
h
Cl
<h
h
h
h
b2
C2
0^2
k.
C2
a^
62
h
h
h
C3
, 2/ =
ag
h
C3
-,2; =
q-3
h
h
(h
h
<h
«i
h.
Cl
a.
h
Cl
(h
h
C2
02
h
^2
a^
h
C2
«3
h
C3
«3
h
C3
«3
h
C3
EXERCISES
1. Evaluate the determinants :
1 2 1
(a)
3 1 3
1 4 1
•
13
id)
4 3
5 -1
2|
1
2
3
(&)
4
5
6
7
8
9
1
(e)
X
1
y
z
1
(0
(/)
-1
1 2
7
3
.
6
-4 9
1
c -h
— c
1 a
h
— c
i|
Ill, §44] SIMULTANEOUS LINEAR EQUATIONS
49
2. Show that
a-\-b
b
3. Evaluate
r^ ''• (a)
b
b + c c
c c + d
\a b c d)
a
b
c
b
c
a
c
a
b
a
a
a
a
a
a
4. Solve by determinants :
2r = 5,
(a) I X— y- z =0,
2z = 1.
(c)
(e)
X— y -
2x-^y-
x + 2y- Sz
x + Sy
2x-6y-l0z
1 1 2
= 7,
= 4,
= -8.
x^
+ 8 = 0,
4 + A-4+ 9 = 0,
(&)
(d)
(/)
(&)
Sx -4y +1 z =8,
2x +3y +Qz =-7,
X - y =4.
a;2 +
3,
i + i-^.+ 15
2X2- y2 4.3 2;2^ 62,
5 iC2 _ 2 2/2 _ 3 ^2 ^ _ 11.
2 3
X— y y -z
4 ^
^ + x X — y
— + ^
W — + X
= 1,
= -7,
= 0.
44. Properties of Determinants. — The advantages of using de-
terminants instead of the longer equivalent algebraic expressions of the
usual kind will be apparent after studying the principal properties of de-
terminants and the geometrical applications that will follow.
(1) A determinant is zero whenever all the elements of any roio, or all
those of any column, are zero.
This follows from the fact that, in the expanded form (§42), every
term contains one element from each row and one from each column.
(2) It follows, for the same reason, that if all elements of any roio {or
of any column) have a factor in common, this factor can be taken out and
placed before the determinant; thus, e.g.,
ai mbi Ci
a2 mbi C2
as mbs Cs
«1
bi ci
a2
bi C2
«3
&3 C3
50
PLANE ANALYTIC GEOMETRY [III, § 44
(3) The value of a determinant is not changed by transposition ; i.e.
by making the columns the rows, and vice versa., preserving their order.
Thus:
ai
bi
Cl
ai a2
as
a2
&2
C2
=
6i 62
bs
as
bz
Cs
Cl C2
Cs
for, by expanding the determinant on the riglit we obtain the same six
terms, with the same signs, as by expanding the determinant on the left.
(4) The interchange of any two rows {or of any two columns) reverses
the sign, but does not change the absolute value, of the determinant.
This also follows directly from the expanded form of the determinant
(§ 42). For, the interchange of two rows is equivalent to interchanging
two subscripts leaving the letters fixed, and this changes every term with
the sign + into a term with the sign — , and vice versa. The interchange
of two columns is equivalent to the interchange of two letters, leaving
the subscripts fixed, which has the same effect.
(5) A determinant in which the elements of any row (column) are equal
to the corresponding elements of any other row {column) is zero.
For, by (4), the sign of the determinant is reversed when any two
rows (columns) are interchanged ; but the interchange of two equal rows
(columns) cannot change the value of the determinant. Hence, denoting
this value by A, we have in this case — A = A, i.e. -4 = 0.
^
EXERCISES
1. Show that
4
6
10
2. Evaluate without expanding ;
-3
4
2 3 4
-1
5
= 2
3 1 5
7
-9
5 7 9
(a)
2
-4 3
7
14 7
, (ft)
4
-8 4
13
11
6
-2
3. Without expanding show that
(c)
1000
4
8
(a)
a b c
d e f
a h i
abc
1 1 1
be a 1
a2 a 1
dbc eca fab
; (b) ca b 1
=
62 b 1
gbc hca iab
ab c 1
c2 c 1
Ill, § 45] SIMULTANEOUS LINEAR EQUATIONS
51
45. Expansion by Minors. The general type of a determinant of
the third order is often written in the form
«11 «12
«13
a21 «22
a23
asi 0532
^33
so that the first subscript indicates the row, the second the column in
which the element stands. Any one of the nine elements is denoted
by ttik.
If in a determinant of the third order, both the row and the column in
which any particular element anc stands be struck out, the remaining ele-
ments form a determinant of the second order, which is called the minor
of the element aik. Thus the minor of a23 is
«ii
«31
«12
«32
r§45
J we have
an
ai2 ai3
021
0522 ^23
«31
«32 ^33
«22
^23
«32 0533
«12 ^13
= «11
+ a2i
+ azi
«32
a33
«12 «13
a22 ^23
an
ai2
a2i
0522
a-ix
^32
az\ azz
an ai3
^21 ^23
= ai2
+ ^22
+ a32
^21 a23
a-ix azz
an ai3
the right-hand member is called the expansion of the determinant by
minors of the (elements of the) first column. It should be noticed, how-
ever, that, while the coeflBcients of an and asi in this expansion are the
minors of these elements, the coefficient of a2i is minus the minor of 021.
The determinant can also be expanded by minors of the second column :
ai3
a23
a33
here the coefficients of ai2 and a32 are minus the minors of these elements
while the coefficient of a22 is the minor of a22 itself. This expansion fol-
lows from the previous one because the value of the determinant merely
changes sign when the first and second columns are interchanged.
Let the student write out the similar development in terms of minors
of the third column.
As the value of the determinant is not changed by transposition (§44
(3)), the detenninant may also be expanded by minors of the elements of
any row.
62
PLANE ANALYTIC GEOMETRY [III, § 46
46. Cofactors. To sum up these results briefly, let us denote by A
the value of the determinant itself, and by Aih the value of the minor of
the element aa-, multiplied hij (— 1)*+*, i.e. the so-called cof actor of a»fc.
We then have :
A = aixAii + a2i^2i + «3i^3i ,
= «12^12 + «22^22 + «32^32 ,
= 0513^13 + «23-423 4- dZzA 33,
and similarly for the expansion by minors, or rather cofactors, of any row.
At the same time it should be noted that if we add the elements of any
column (row) each multiplied by the cofactors of any other column (row),
the result is always zero. Thus it is readily verified that
«11^12 + «21-422 + «31-<432 = 0,
«11^13 + «21^23 + «31^33 = 0,
auAn + «22^21 + «32^31 = 0,
etc. This property was used in § 4L
47. Sum of Two Determinants, if all the elements of any
column (or row) are sums, the determinant can be resolved into a sum of
determinants. Thus, if all elements of the firs: column are sums of two
terms, we find, expanding by minors of the first column :
a2+wi2
a3+wi3
(«i + wii)
+ (a2 + wt2)
+ (a3+m3)
=
ai
b% c^
&3 C3
+ 052
bz C3
bi ci
+ az
bi ci
bi Ci
+ mi
62 C2
63 C3
&1 Ci
bi Ci
02 Ca
ai 61 ci
Wi 61 Ci
=
a2 62 <h
+
ma 62 C2
'
Oz
63 c
3
wis 63
Cs
Let the student show, by interchanging rows and columns, that the
same property holds for rows.
As any row (column) can be made the first by interchanging it with the
first and changing the sign of the determinant, this decomposition into
the sum of two determinants is possible whenever every element of any
one row or column is a sum.
Ill, § 47] SIMULTANEOUS LINEAR EQUATIONS
53
As a particular case we have
ai-^bi 61 ci
ai &i ci
61 61 Ci
«! 61 Ci
a2 + &2 62 C2
=
a2 &2 C2
+
&2 &2 C2
=
(Z2 62 C2
as + bs 63 ^^3
^3 h C3
63 &3 C3
053 bs C3
since the second determinant, which has two equal columns, is zero by
(5), § 44. We conclude that the value of a determinant is not changed
by adding to each element of any row {column) the corresponding element
of any other row (column). Indeed, owing to (2), § 44, we can add to
each element of any row (column) the corresponding element of any
other row (column) multiplied by one and the same factor. This property
is of great help in reducing a given determinant to a more simple form
and evaluating it.
In the case of a numerical determinant, it is often best after taking out
the common factors from any row or column to reduce two elements of
some row or column to zero, by addition or subtraction. Thus, taking
out the factors 2 from the third column and 3 from the second row,
we have
2
3
-14
2
3
-7
3
18
-12
= 6
1
-2
4
8
18
-4
8
9
subtracting twice the second row from the first and adding 4 times the
second row to the third, we find
A = 6
-9
-3
-9
-3
1
6
-2
= -« S2
J =18
32
1
= -622.
EXERCISES
1. Evaluate the determinants :
(a)
(d)
1 3 7
3 5 9
,
4 8 16
6 33 9
14 21 35
26 39 4
12 1
(ft)
(e)
27
26
27
31
33
36
43
44
45
7
17
29
11
19
31
13
23
37
He)
(/)
17
34
51
28
72
38
?
39
65
52
2
-3
40
5
7
-10
3
-2
(
30l
\b
54
PLANE ANALYTIC GEOMETRY [III, § 47
2. 8how that
b + c a 1
c -h a b 1
a+ feci
(a)
= 0;
(&)
(c)
(d)
6i + ci
62 + C2
63 + C3
a-hb
a + 2b
a + Sb
ci + ai
C2+ a2
C3+ as
a + 46
a + 5b
a + 66
1 a'2
-ff^
a3
-#
1
a2
a3
1 &2 _ d2 b^
-d^
=
1
62
63
1 c2-d-^ c3
-#
1
C2
C3
ai + &i
«i
61 Ci
a2 + 62
— 2
a2
62 C2
>
a3 + 63
as
&3 C3
a + 76
a + 86
= 0.
a + 9b
48. Elimination. Three hoynogeneous linear equations,
(3)
a^x + 61?/ + CiZ = 0,
a^ -f &2^ + C22; = 0,
«'3« + 632/ + C32 = 0,
are obviously satisfied by x = 0, y = 0, z=0. Can they have
other solutions?
Solving the equations by the method of § 43, and denoting
the determinant of the equations for the sake of brevity by A,
we find since Aij = 0, fcg = 0, A^a = :
Ax=0, Ay = 0, Az = 0.
Hence, if x, y, z are not all three zero, we must have ^ = 0.
Three homogeneous linear equations can therefore have solutions
that are not all zero only if the determinant of the equations is
equal to zero.
If X, for instance, is different from zero, we can divide each
of the three equations by x and then eliminate y/x and z/x be-
tween the three equations. The result is ^ = 0, i.e.
tti hi Ci
a^ 62 C2 =0.
Ctg 63 Ci
Ill, § 49] SIMULTANEOUS LINEAR EQUATIONS 55
Thus, the result of eliminating the three variables betiveen three
homogeneous linear equations is the determinant of the equations
equated to zero. (Compare § 40.)
Solving the first and second equations for y/x, z/Xy we obtain
X
y
z
bi Ci
c, a,
a, 61
62 C2
C2 0^2
(X2 62
provided the denominators are all different from zero.
With the notation of § 46, this can be written x:y : z=Azi : ^32 : ^33-
If we solve the third and first or second and third equations for y/x, z/x^
we find, respectively, x :y: z = A21 : A22 '■ A23, or x :y :z = An '• ^12 : ^i3-
Hence, whenever ^ = 0, we can find the ratios of the variables unless all
the minors of A are zero.
49. Geometric Applications. The equation of a line
through two points Pj {x-^, 2/1) and Pg fe 2/2) can be found as fol-
lows. The equation of any line must be of the form (§ 30)
(4) Ax-irBy+C=^0.
The question is to determine the coefficients A, B, C, so that
the line shall pass through the points P^ and Pg. If the line is
to pass through the point P^, the equation must be satisfied by
the coordinates x^, y^ of this point, i.e. we must have
^1 + ^^1+0 = 0;
this is the first condition to be satisfied by the coefficients. In
the same way we find the second condition
Ax^-\- By,^^C=0.
We might calculate from these two conditions the values of A/C
and B/G and then substitute these values in the first equation.
But as this means merely eliminating A, B, C between the
three equations, we can obtain the result directly (§ 48) by
equating to zero the determinant of the coefficients of A, B, C.
56
PLANE ANALYTIC GEOMETRY [III, § 49
Thus the equation of the line through two points P^, P^ is
(5)
X
y
1
«!
2/1
1
X^
2/2
1
= 0.
Observe that this equation is evidently satisfied if x^ y are re-
placed either by x^, i/i or by x^, 2/2 (see (5), § 44).
50. Area of a Triangle. The area ^ of a triangle PiPoP^
in terms of the coordinates of its vertices Pi(xij 2/i)> ^^2(^2? 2/2)?
^3(3^3,2/3)13:
for, upon expanding this determinant, we find the value given
before in § 14.
It will now be seen that the determinant equation (5) of the
line through two points given in § 49 merely expresses the
fact that any point (x,'y) of the line forms with the given
points (xi, 2/1) and (x^, y^ a triangle whose area is zero.
■A = \
X,
2/1
1
x^
2/2
1
X,
Vs
1
EXERCISES
1. Write down the equation of the line through (2, 3), (—2, \); ex-
pand the determinant by minors of the first row ; determine the slope and
the intercepts ; sketch the line.
2. Find the equation of the line through the points : (3, — 4) and
(0, 2) ; (0, 6) and (a, 0); (0, 0) and (2, 1).
3. Find the area of the triangle whose vertices are the points (1, 1),
(2, -3), (5, -8).
4. Find the area of the quadrilateral whose vertices are the points
(3, -2), (4, -5), (-3,1), (0,0).
6. If the base of a triangle joins the points (— 1, 2) and (4, 3), on
what line does the vertex lie if the area of the triangle is equal to 6 ?
Ill, §50] SIMULTANEOUS LINEAR EQUATIONS
57
6. Find the coordinates of the common vertex of the two triangles of
equal area 3, whose bases join the points (3, 5), (6, — 8) and (3, — 1),
(2, 2), respectively.
7. Show that the area of any triangle is four times the area of the
triangle formed by joining the midpoints of its sides.
8. Show that the sum of the areas of the triangles whose vertices are
(a, d), (2 6, c), (6 c, f), and (Sa,d), (4&, e), (3 c,/) is given by the
determinant
2a d 1
Bb e 1
4c / 1
9. Show that the lines joining the midpoints of the sides of any
triangle divide the triangle into four equal triangles.
10. Show that the condition that the three lines Ax -{- By ■{- C = 0^
A'x + B'y + C" = 0, A"x + B"y + C" = meet at a point is
ABC
A' B' C =0.
A" B" C"
11. Show that the straight lines Sx + y — l=0, x—Sy-\-lS = 0,
2x— y-{-6=^0 have a common point.
12. For what values of s do the following lines meet in a point :
4x-Qy-\-s = 0, sx-S6y = 0,x-{-y-l=0?
13. Show that the altitudes of any triangle meet in a point.
14. Show that the medians of any triangle meet in a point.
15. Show that the line through the origin perpendicular to the line
through the points (a, 0) and (0, b) meets the lines through the points
(a, 0), (— &, &) and (0, 6), (a, — a) in a common point.
16. Show that the distance of the point Pi(xi, y{) from the line joining
the points P2(,X2, 1/2) and Pz^xs, ys) is
xi y\
1
X2 yt
1
xz yz
1
y/ixz-x.y^+ (2/3-2^2)2
CHAPTER IV
RELATIONS BETWEEN TWO OR MORE LINES
51. Angle between Two Lines. We shall understand by
the angle (I, V) = 6 between two lines I and I' the least angle
through which I must be turned coun-
terclockwise about the point of inter-
section to come to coincidence with l'.
This angle is equal to the differ-
ence of the slope angles a, a' (Fig. 27)
of the two lines. Thus, if a' > a, we
have B= a! — a, since a' is the exterior
angle of a triangle, two of whose interior angles are a and 6.
It follows that
tan a! — tan a ■
Fig. 27
(1)
tan Q — tan {a! — a)
1 -f- tan a tan a!
If the equations of I and X are
y = mx -\- b, y = m'x + 6',
respectively, we have tan a = m, tan a' = m' ; hence
m' — m
(2)
tan^ =
1 4- mm'
If the equations of I and I' are
^a; + 52/ -f C =0,
respectively, we have tan a = — ^1/-B, tan a' = — ^'/^' 5 hence
AB' - AB
(3)
tan^ =
AA'+BB^'
58
IV, § 52] RELATIONS BETWEEN LINES 59
52. It follows, in particular, that the two lines Z and Z', § 51,
are parallel if and only if
m' = m, or AB' - A'B = ;
and they are perpendicular to each other if and only if
m' = --, ovAA' + BB'=0.
m
(Compare §§ 27, 31.) Hence, to write down the equation of
a line parallel to a given line, replace the constant term by an
arbitrary constant ; to write down the equation of a line per-
pendicular to a given line, interchange the coefficients of x and
y, changing the sign of one of them, and replace the constant
term by an arbitrary constant.
EXERCISES
1. Determine whether the following pairs of lines are parallel or per-
pendicular : 3x + 2?/ — 6 = 0, 2x-3?/ + 4=z0; 5ic + 3y-6=0,
10x + 6y4-2 = 0;2x-|-5y-14=0, 8x-3?/ + 6=0.
2. Find the point of intersection of the Hne 5x + 8y + 17=0 with its
perpendicular through the origin,
3. Find the point of intersection of the lines through the points (6, —2)
and (0, 2), and (4, 5) and (-1,-4).
4. Find the perpendicular bisector of the line-segment joining the
point (3, 4) to the point of intersection of the lines 2x — y + 1 = Q and
3 X 4- 2/ - 16 = 0.
6. Find the lines through the point of intersection of the lines 5 x— z/ =0,
x + 7i/ — 9 = and perpendicular to them.
6. Find the area of the triangle formed by the lines 3 x + 4 y = 8,
6 X — 5 2/ = 30, and x = 0.
7. Find the area of the triangle formed by the lines x + y — 1 = 0,
2 X + y + 5 = 0, and X - 2 !/ - 10 = 0.
8. Find the point of intersection of the lines
(a) ^ + f=l, f + I=l.
ah ha
(h) - + |=1, y = mx-\-h-
a
60 PLANE ANALYTIC GEOMETRY [IV, § 52
9. Find the area of the triangle formed by the lines y = miX + 6i,
y = m2X + &2 and the axis Ox.
10. The vertices of atriangle are (5, — 4), (— 3, 2), (7,6). Find the
equations of the medians and their point of intersection.
11. Find the angle between the lines 4 x—S y—G=0 and x—7 y-\-Q=0.
12. Find the tangent of the angle between the lines (a) 4 x—Sy-\-6=0
and9a; + 22/-8 = 0; (b) 3a; + 6y-ll=0 and x-\-2y-S = 0.
13. Find the two lines through the point (6, 10) inclined at 45° to
the line 3a;-2?/-12=:0.
14. Find the lines through the point (— 3, 7) such that the tangent of
the angle between each of these lines and the line 6.x — 2i/ + ll = 0isJ.
15. Show that the angle between the lines Jtc + J5y + C = and
(A + B)x -{A- B)y + D = is 45°.
16. Find 'the lines which make an angle of 45° with the line
4x — 7y + 6=0 and bisect the portion of it intercepted by the axes.
17. The hypotenuse of an isosceles right-angled triangle lies on the line
Sx — 6y-n==0. The origin is one vertex ; what are the others ?
53. Polar Equation of Line. The position of a line in the
plane is fully determined by the length p = ON (Fig. 28) of the
perpendicular let fall from the origin on
the line and the angle /3 = xON made by
this perpendicular with the axis Ox.
Then p and /8 are evidently the polm^
coordinates of the point -^ (§ 16). Let
P be any point of the line and OP = r,
xOP— d> its polar coordinates. As the
Fig *^8
projection of OP on the perpendicular
ON is equal to ON, and the angle NOP = <^ — ft we have
(4) rGO{i(<f> — p)=p.
This is the equation of the line NP in polar coordinates.
IV, §54] TWO OR MORE LINES 61
54. Normal Form. The last equation can be transformed to
Cartesian coordinates by expanding the cosine :
r cos <^ cos P + r sin <f>sm p=p
and observing (§ 17) that r cos <l> = x, r sin <i> = y\ the equation
then becomes
(5) ai^cosp+ y8inp=ip.
This equation, which is called the normal form of the equation
of the line, can be read off directly from the figure ; it means
that the sum of the projections of x and y on the perpendicular
to the line is equal to the projection of r (§ 20).
Observe that in the normal form (5) the number p is always
positive, being the distance of the line from the origin, or the
radius vector of the point JSf. Hence x cos ^ -f- y sin ^ is always
positive ; this also appears by considering that x cos /3 -\-y sin ^
is the projection of the radius vector OP on ON, and that this
radius vector makes with ON an angle that cannot be greater
than a right angle.
The angle l3 = xONiSj as a polar angle (§ 16), always under-
stood to be the angle through which the axis Ox must be turned
counterclockwise about the origin to make it coincide with ON;
it can therefore have any value from to 2 tt. By drawing the
parallel to the line NP through the origin it is readily seen
that, if a is the slope angle of the line NP, we have
according as the line lies on one side of the origin or the other,
angles differing by 2 tt being regarded as equivalent. Thus, in
Fig. 28, « = 120°, /? = «+! 7r = 120°+ 270° = 390°, which is
equivalent to 30°. For a parallel on the opposite side of the
origin we should have ^ = «-}- 1 tt = 120° + 90° = 210°.
62 PLANE ANALYTIC GEOMETRY [IV, § 55
55. Reduction to Normal Form. The equation
Ax-\-By^C=0
is in general not of the form (5), since in the latter equation
the coefficients of x and y, being the cosine and sine of an
angle, have the property that the sum of their squares is equal
to 1, while in the former equation the sum of the squares of
A and B is in general not equal to 1. But the general equation
Ax-{-By-\-C=0
can be reduced to the normal form (5) by multiplying it by
a factor k properly chosen ; we know (§ 30) that the equation
hAx-\-'kBy^'kG=0
represents the same line as does the equation Ax-\-By-\-G=0.
Now if we select k so that
kA = cos p, kB = sin /3, kC = — 1>,
the equation Ax + By-\-C=0 reduces to the normal form
xQos p + y sin p — 2) = 0. The first two conditions give
A;M2 + k'^B' = cos2 /3 + sin^ ^ = 1,
whence A; = ±
VA^-hB"
Since the right-hand member p in the normal form (5) is posi-
tive, the sign of the square root must be selected so that kC
becomes negative. We have therefore the rule :
\ To reduce the general equation Ax -\-By-{-C =0 to the normal
J form
\ ajcos^ + 2/ sin/3 — j9= 0,
/ divide by — ■\/ A? + B^ when C is positive and by -^^A^-\-B^
\wjien C is negative.
Then the coefficients of x and y will be cos ft sin ft respec-
tively, and the constant term will be the distance p of the line
from the origin.
IV, § 56] TWO OR MORE LINES 63
Thus, to reduce 3a; + 22/H-5 = 0to the normal form, divide
by _ V3'' + 22 = - Vl3 ; this gives
3 . ^ 2 • 5
cos B = , sm « = Tzn, —p = 1= ;
VI3 V13 V13
i.e. the normal form is
3 2 5
7=^ ;=2/ =
V13 Vl3 V13
The perpendicular to the line from the origin has the length
5/ Vl3 ; and as both cos ^ and sin ft are negative, this perpen-
dicular lies in the third quadrant. Draw the line.
Reduce the equation 3 a; + 2?/ — 5 = to the normal form.
^ 56. Distance of a Point from a Line. If, in Fig. 28, we
take instead of a point P on the line any point Pi {x^, 2/1)
not on the line (Fig. 29), the expression \ ^
Xy cos P + Vi sin j8 is still the projection on
ON (produced if necessary) of the radius
vector OPi. But this projection OS differs
from the normal ON = p to the line. The
figure shows that the difference ' 1 \ , ~
Xy cos )8 + 2/1 sin y8 — p = OaS — 0N= N8 fig. 29 ^-
is equal to the distance N^P^ of the point Pj from the line.
Thus, to find the distance of any point Pj (x^, 2/1) from a line
whose equation is given in the normal form
a; cos /8 + 2/ sin ^ — p = 0,
it sufiices to substitute in the left-hand member of this equa-
tion for X, y the coordinates x^, 1/1 of ^^^ point P^. The expression
iCi cos /? -f 2/1 sin ^ — p
then represents the distance of P^ from the line.
If this expression is negative, the point P^ lies on the same
side of the line as does the origin ; if it is positive, the point
64
PLANE ANALYTIC GEOMETRY
[IV, § 56
Pi lies on the opposite side of the line. Any line thus divides the
plane into two regions which we may call the positive and nega-
tive regions ; that in which the origin lies is the negative region.
To find the distance of a point Pi (x^, y{) from a line given in
the general form
Ax-\-Bi/-i-C=0,
we have only to reduce the equation to the normal form (§ 55)
and then apply the rule given above. Thus the distance is
Ax, + By, + C ^^ Ax,-^By,-\-C ^
- V^-P + B" V^2-f^
according as C is positive or negative.
57. Bisector of an Angle. To find the bisectors of the
angles between two lines given in the normal form
x cos /8 4- 2/ sin ^—p=zO,
X cos /?' + y «iii P' —p' = 0,
observe that for any point on either bisector its distances from
the two lines must be equal in absolute value. Hence the
equations of the bisectors are
a; cos ^ + ?/ sin )8 — p = ± (a; cos y8' + 2/ sin yS' — i>').
To distinguish the two bisectors, ob-
serve that for the bisector of that pair
of vertical angles which contains the
origin (Fig. 30) the perpendicular dis-
tances are, in one angle both positive,
in the other both negative ; hence the
plus sign gives this bisector.
If the equations of the lines are
given in the general form
Ax + By + C = 0, A'x -f- B'y + C = 0,
first reduce the equations to the normal form, and then apply
the previous rule.
Fig. 30
IV, §57] TWO OR MORE LINES 65
EXERCISES
1. Draw the lines represented by the following equations :
(a) rcos(0-^7r)=6.
(e) r cos (0 + f tt) = 3.
(6) r cos (0 - tt) = 4.
(/) rsin (0-i7r) =8.
(c) r cos = 10.
(g) rsin (0 + |^) = 7.
((?) r sin = 5.
(A) r cos (0 - 1 tt) = 0.
2. In polar coordinates, find the equations of the lines : (a) parallel to
and at the distance 5 from the polar axis (above and below) ; (b) per-
pendicular to the polar axis and at the distance 4 from the pole (to the
right and left) ; (c) inclined at an angle of |ir to the polar axis and at
the distance 12 from the pole.
3. Express in polar coordinates the sides of the rectangle OABG if
OA = 6 and AB = 9, OA being taken as polar axis.
4. What lines are represented by (5) when p is constant, while /3
varies from zero to 2 ir ? What lines when p varies while j3 remains con-
stant ?
5. The perpendicular from the origin to a line is 5 units in length and
makes an angle tan-i y\- with the axis Ox. Find the equation of the line.
6. Reduce the equations of Ex. 8, p. 34, to the normal form (5),
7. Find the equations of the lines whose slope angle is 150° and which
are at the distance 4 from the origin.
8. What is the equation of the line through the point ( — 3, 6) whose
perpendicular from the origin makes an angle of 120^ with the axis Ox ?
9. For the line 7a;— 24?/ — 20 =0 find the intercepts, slope, length
of perpendicular from the origin and the sine and cosine of the angle
which this perpendicular makes with the axis Ox.
10. Find by means of sin ^3 and cos ^ the quadrants crossed by the line
4x — 5y = S.
11. Put the following equations in the form (5) and thus find p, sin /3,
cos /3:
(a)y=mx-\-b. (b) - +^ = 1. (c)3« = 4y.
a b
12. Is the point (3, — 4) on the positive or negative side of the line
through the points (— 5, 2) and (4, 7) ?
66 PLANE ANALYTIC GEOMETRY [IV, § 57
13. Is the point (—1, — f ) on the positive or negative side of the line
4x-9y-S = 0?
14. Find by means of an altitude and a side the area of the triangle
formed by the lines 3a5 + 2i/ + 10 = 0, 4x-3?/+16 = 0, 2cc + ?/-4
= 0. Check the result with another altitude and side.
15. Find the distance between the parallel lines (a) Hx— 6y— 4 =
and 6 X - 10 y + 7 = ; {h) 5 x + 7 y + 9 = and 15 x + 21 y - 3 = 0.
16. What is the length of the perpendicular from the origin to the line
through the point (—5, — 4) whose slope angle is 60" ?
17. What are the equations of the lines whose distances from the
origin are 6 units each and whose slopes are | ?
18. Find the points on the axis Ox whose perpendicular distances from
the line 24 x ■- 7 ?/ — 16 = are ±5.
19. Find the point equidistant from the points (4, — 3) and (—2, 1),
and at the distance 4 from the line 3x — 4?/ — 5 = 0.
20. Find the line parallel to 12 x — 5?/ — 6 = and at the same distance
from the origin ; farther from the origin by a distance 3.
21. Find the two lines through the point (1, -y^) such that the perpen-
diculars let fall from the point (6, 5) are of length 5.
22. Find the line perpendicular to4x — 7«/ — 10 = which crosses the
axis Ox at a distance 6 from the point (— 2, 0).
23. Find the bisectors of the angles between the lines: (a) x—y —4=0
and 3 X + 3 y + 7 = ; (6) 6x - 12 y - 16 = and 24 x + 7y + 60 = 0.
24. Find the bisectors of the angles of the triangle formed by the lines
5 X + 12 y + 20 = 0, 4 X — 3 2/ - 6 = 0, 3 X - 4 y + 5 = and the centerof
the circle inscribed in the triangle.
25. Find the bisector of that angle between the lines 3 x — VS ?/+ 10=0,
V2 x + y — 6 = 0in which the origin lies.
26. If two lines are given in the normal form, what is represented by
their sum and what by their difference ?
27. Show that the angle between the lines x + y = and x — y = is
90° whether the axes are rectangular or oblique.
IV, § 58] TWO OR MORE LINES 67
58. Pencils of Lines. All lines through one and the same
point are said to form a pencil; the point is called the center of
the pencil. If
^^ \A'x + B'y-^C'=:0
are any two differeijjt-iilies of a pencil, the equation
(7) Ax-\-By+C+k(A'x-hB'y-]-C')=0,
where k is any constant, represents a line of the pencil. For,
the equation (7) is of the first degree in x and y, and the coeffi-
cients of X and y cannot be both zero, since this would mean
that the lines (6) are parallel. Moreover, the line (7) passes
through the center of the pencil (6) because the coordinates of
the point that satisfies each of the equations (6) also satisfy
the equation (7).
All lines parallel to the same direction are said to form a
pencil of parallels. It is readily seen that if the lines (6) are
parallel, the equation (7) represents a line parallel to them.
EXERCISES
1. Find the line: (a) through the point of intersection of the lines
4 ic — 7 y + 5 =0, 6aj + 11 y — 7=0 and the origin ; (6) through the
point of intersection of the lines 4a; — 2y — 3 = 0, x^-y — 5 = and
the point (—2, 3) ; (c) through the p^nt of intersection of the lines
4ic — 5?/ + 6 = 0, y — x — S = 0, of slope 3 ; (d) through the intersection
of5x — 62/4-10 = 0, 2x + 3y— 12 = 0, perpendicular to 4 y + a; = 0.
2. Find the line of the pencil x— 5 = 0, y -\-2 = that is inclined to
the axis Ox at 30°.
3. Determine the constant b of the line y = 3x+ b so that this line
shall belong to the pencil Sx — iy + 6 = 0, x = 6.
4. Find the line joining the centers of the pencils x — Sy = 12,
5x— 2y = 1 and x-{-y = 6, 4tx — 5y = S.
5. Find the line of the pencil 4x-5y-12 = 0, 3a; + 22/-16=0
that makes equal intercepts on the axes.
68 PLANE ANALYTIC GEOMETRY [IV, § 59
69. Non-linear Equations representing Lines. When two
lines are given, say
Ax-{-By-\-C=0,
then the equation
{Ax -f JB2/ + C){A'x + By + 0') = 0,
obtained by multiplying the left-hand members (the right-hand
members being reduced to zero) is satisfied by all the points
of the first given line as well as all the points of the second
given line, and by no other points.
The product equation which is of the second degree is there-
fore said to represent the two given lines. Similarly, by equat-
ing to zero the product of the left-hand members of the equations
of three or more straight lines (whose right-hand members are
zero) we find a single equation representing all these lines.
An equation of the 7ith degree may therefore represent n
straight lines, viz. when its left-hand member (the right-hand
member being zero) can be resolved into n linear factors, with
real coefficients.
EXERCISES
1. Find the common equation of the two axes of coordinates.
2. Show that n lines through the origin are represented by a homo-
geneous equation (i.e. one in \^ich all terms are of the same degree in
X and y) of the nth degree.
3. Draw the lines represented by the following equations :
(a) (x -a)(y-b)= 0. (/) xy - ax = 0.
(6) 3x^-xy-4y'^ = 0. (g) y^ - ^y^ ■¥ Qy = 0.
(c) rK2 _ 9 1/2 = 0. (h) yfiy-xy = 0.
{d) ax"^ + 6^2 = 0. (0 y^-Q xy"^ + 11 x^y - 6 a;^ = 0.
(e) a:2 - iK - 12 = 0.
4. What relation must hold between a, h, b, if the lines represented
by ax^ -\-2hxy + by^ = are to be real and distinct, coincident, imag-
inary ? •
IV, § 59] TWO OR MORE LINES 69
MISCELLANEOUS EXERCISES
1. Find the angle between the lines represented by the equation
ayi^ + 2 hxy + hy'^ — 0. What is the condition for these lines to be per-
pendicular ? coincident ?
2. Reduce the general equation Ax -\- By -{- C = to the normal
form xoos p + y sin j3 = p by considering that, if both equations represent
the same line, the intercepts must be the same.
3. Find the line through (xi , yi) making equal intercepts on the axes.
4. Find the area of the triangle formed by the hues y = miX + 6i ,
y = m2X -\- b2 >, y = b.
5. What does the equation = const, represent in polar coordinates ?
6. Find the polar equation of the line through (6, v) and (4, | nr).
7. Derive the determinant expression for the area of a triangle (§14)
by multipljdng one side by half the altitude.
8. The weights lo, W being suspended at distances d, Z), respectively,
from the fulcrum of a lever, we have by the law of the Jever WD = icd.
If the weights are shifted along the lever, then to every value of d cor-
responds a definite value of D ; i.e. i> is a function of d. Represent this
function graphically ; interpret the part of the line in the third quadrant.
9. A train, after leaving the station yl, attains in the first 6 minutes,
li miles from A, the speed of 30 miles per hour with which it goes on.
How far from A will it be 50 minutes after starting? (Compare Ex-
ample 4, § 29.) Illustrate graphically, taking s in miles, t in minutes.
10. A train leaves Petroit at 8 hr. 25 m. a.m. and reaches Chicago at
4 hr. 5 m. p.m. ; another train leaves Chicago at 10 hr. 30 m. a.m. and
arrives in Detroit at 5 hr. 30 m. p.m. The distance is 284 miles. Regard-
ing the motion as uniform and neglecting the stops, find graphically and
analytically where and when the trains meet. If the scale of distances
(in miles) be taken 1/20 of the scale of times (in hours), how can the
velocities be found from the slopes ?
11. A stone is dropped from a balloon ascending vertically at the rate
of 24 ft. /sec; express the velocity as a function of the time (Example 5,
§ 29) . What is the velocity after 4 sec. ?
12. How long will a ball rise if thrown vertically upward with an
initial velocity of 100 ft. /sec. ?
CHAPTER V
PERMUTATIONS AND COMBINATIONS. DETERMI-
NANTS OF ANY ORDER
60. Introduction. In using determinants of the second and
third order we have seen how advantageous it is to arrange
conveniently the symbols of an algebraic expression. Before
proceeding to the study of the general determinant of the wth
order, we must discuss very briefly that branch of algebra
which is concerned with the theory of arrangements and
changes of arrangement (permutations and combinations).
The results are important not only for determinants, but are
used very often, even in the common affairs of life ; they form,
moreover, the basis of the theory of " choice and chance," or of
probabilities.
The " things " to be arranged or combined need not be num-
bers (as they are in a determinant), but may be any what-
ever, provided they are, and remain, clearly distinguishable
from each other ; we shall call them elements and designate
them by letters a, h, c, etc.
61. Permutations. Any two elements, a and 6, can obvi-
ously be arranged in a row in 2 ways :
ah, ha.
Three elements a, 5, c can be arranged in a row in 6, and only
6, ways:
ahc hac cab
acb hca cha
The question arises: in how many ways can ^i elements be
arranged in a row ?
70
V, §62] PERMUTATIONS AND COMBINATIONS 71
Any arrangement of n elements in a row is called a permu-
tation. It is found by trial that the number of permutations of
n elements increases very rapidly with their number n. Thus
for 4 elements it is 24, for 5 elements 120. It will be shown
that for n elements the number of permutations zs 1 • 2 • 3 • • • n.
This expression, the product of the first n positive integers, is
briefly designated by n !, or \n, and is called factorial n :
n! = l -2 .3...%.
If we denote by P„ the number of permutations of n ele-
ments our proposition is
62. Mathematical Induction. The proof of the proposition
that P^ = nl is obtained by an important method of reasoning
called mathematical induction.
By actual trial we can readily find that P^ = 1, Pg = 2,
Pg = 6, and with sufficient patience we might even ascertain
that Pe = 720. But to prove the general proposition that
P^ = 7i! we must look into the method by which in the
particular cases we make sure that we have found all the pos-
sible permutations. This method consists in proceeding step
by step :
Seeing that 2 elements have 2 permutations, we form the
permutations of 3 elements by taking each of the 3 elements
and associating with it the 2 permutations of the remaining
two ; we thus find that Pg = 3 • 2 = 6.
Similarly, to form the permutations of 4 elements we asso-
ciate each of the 4 with the 6 permutations of the remaining 3 ;
this gives P4 = 4 .3! = 4!
This leads us to expect that P^ = nl The actual proof rests
on two facts : (a) the special fact, found by actual trial, that
72 PLANE ANALYTIC GEOMETRY [V, §62
e.g. P2 = 2 ! ; (6) the general law that the number of permuta-
tions of n-\-l elements is found by associating each of the
n -f 1 elements with the P„ permutations of the remaining 71,
i.e. that
-p„+.=(»+i)^„-
Knowing from (a) that P2 = 2 ! we find from this formula that
P3 = 3 . P2 = 3 . 2 ! = 3 ! ; in the same way that P4 = 4 . 3 ! = 4 !
etc.
Notice that mathematical induction is not merely a method
of trial and experiment. It requires that we should know not
only one special case of the general formula to be proved, but
also the law by which we can proceed from every special case to
the next, i.e. from n to ?i + 1 whatever the value of n. This law
is a result, not of trial or induction, but of deductive reasoning.
In our case it is expressed by the formula P„+i = (n -{- 1)P„.
The method of mathematical induction is therefore often called
reasoning from n to n-\-l.
63. Permutations by Groups. A somewhat more general
problem in permutations is suggested by the following exam-
ple: In an office there are two vacancies, one at $1000, the
other at $800. There are 5 applicants for either of the 2
positions ; in how many ways can the positions be filled ?
The first vacancy can be filled in 5 ways, and then the sec-
ond can still be filled in 4 ways ; hence there are 5 • 4 = 20
ways. Denoting the applicants by a, &, c, d, e the 20 possi-
bilities are :
ab
ac
ad
ae
ha
be
bd
be
ca
cb
cd
ce
da
db
do
de
ea
eb
ec
ed
V, §64] PERMUTATIONS AND COMBINATIONS 73
The general problem here suggested is that of finding the
number of permutations of n elements k at a time, where Tc <n.
Each permutation here contains k elements ; and we have to
fill the k places in all possible ways from the n given elements.
The first place can be filled in n ways. The second can then be
filled in ?i — 1 ways ; hence the first and second places can be
filled in n(ii — 1) ways. The third place, when the first two are
filled, can still be filled in n — 2 ways, so that the first three
places can be filled in n{n — V)(ji — 2) ways. Proceeding in this
way we find that the k places can be filled in ti (n — 1) (n — 2) •••
(n -~k-{-l) ways.
Thus the number of permutations of n elements, A; at a time,
which is denoted by „P;t, is
„P, = n{n - l){n _ 2) ... (n - fc + 1).
Notice that in ^P^ there are as many factors as places to be
filled, viz. k ; the first factor being n, the second n — 1, etc., the
A:th will be n — {k -- 1) = n — k -\-l.
lik^nwQ have the case of § 61 ; i.e. „P„ == P^.
As ?i! = n(?i — 1) ... (n-'k-\-l) • {n — k){7i — k — l) ..-2.1
= n{n — 1 ) '" {n — k-\-l) • {n — k)\, the expression for „P^ can
also be written in the form
p _ n!
{n-k)\
64. Combinations. If, in the problem of § 63, the 2
vacancies to be filled are positions of the same rank (as to
salary, qualifications required, etc.), the answer will be differ-
ent. We have now merely to select in all possible ways 2 out
of 5 applicants, the arrangements ah and ha, ac and ca, etc.,
being now equivalent. Therefore the answer is now 20 divided
by 2, i.e. 10, as can readily be verified directly : ah, ac, ad, oe,
be, bdj be, cd, ce, de.
74 PLANE ANALYTIC GEOMETRY [V, § 64
If there were 3 vacancies, the number of ways of filling
them from 4 applicants, when the positions are different, is
4P3 = 4 . 3 • 2 = 24 ; but when the positions are alike, the
number is 24 divided by the number of permutations of 3
things, i.e. 24/6 = 4.
A set of k elements selected out of n, when the arrangement
of the k elements in each set is indifferent, is called a combina-
tion. The number of combinations of k elements that can be
selected from n elements is denoted by ^C^ ; to find this num-
ber we may first form the number ^P^ of permutations of n
elements A; at a time, and then divide by the number Pj^=ik\
of permutations of k elements. Thus
p _ n{n — l) ••' {n — k-\-l) _ n\
" *~ 1.2 ...A; ~k\{n-k)\'
The number of combinations of n elements that can be
selected from n elements is clearly 1 ; indeed, for A: = n our
first expression gives ^(7„ = 1.
EXERCISES
1. Find the value of n if ^ *
(a) ^ = 5. ^ , {hy ^■= 20. (c) p. = 40320.
2. Show that
(a) nGk = nGn-lc> (&) nCk"^ nCk-^= n+lCk. (c) A;n+lC*= (n + 1)„C*-1.
3. Prove by mathematical Induction that :
(a) 1 + 2 + 3 + - + n = I n(n + 1).
(5) 12 + 22 + 3'-'+ ••• +n2 = ^n(n+l)(2n + l).
(c) 13 + 23 + 33+ ... +n3=[^n(n + l)]2 = (l +2 + 3+ •.• + n)2.
{d) 1 + 3 + 5 + ... +(2 n - 1)= n2.
(c) 2 + 4 + 6 + ... + 2 n = n{n + 1).
(/) 1.2 + 2.3 + 3.4+ ... +n(7i + l)=in(w + l)(n + 2).
(9') T~^ + ^7—^ + ^7— ; + ••• +
1.22.33.4 n{n + \) n + \
V, §65] PERMUTATIONS AND COMBINATIONS 75
4. A pile of shot forms a pyramid with n shot on a side at the base.
How many shot in the pile if the base is a square ? an equilateral triangle ?
6. Three football teams plan a series of games so that each team will
play the other two teams 4 times. How many games in the schedule ?
6. In how many ways can a committee of 3 freshmen and 2 sopho-
mores be chosen from 8 freshmen and 5 sophomores ?
7. In how many ways can the letters of the word equal be arranged
in a row four letters at a time ?
8. From the 26 letters of the alphabet, in how many ways can four
different letters, one of which is d, be arranged in a row ?
9. How many numbers of three digits each can be formed with 1,
2, 3, 4, 5, no digit being repeated ? How many of these numbers are
even ? odd ?
10. From a company of 60 men, how many guards of 4 men can be
formed? How many times will one man (A) serve ? How many times
will A and B serve together ?
11. Which is the largest of the numbers „(7i, „C2, nOs, ••• „0„_i,
when n is even ? odd ?
12. How many straight lines are determined by 12 points, no 3 of
which are in a line ?
13. How many triangles are determined by 10 points, no 3 of which
are in a line ?
65. Inversions in Permutations. When n elements ai, ag,
as, ••• a„, distmguished by their subscripts, are given, their arrange-
ment, with the subscripts in the natural order of increasing numbers,
is called the principal permutation. In every other permutation of these
elements it will occur that lower subscripts are preceded by higher ones.
Every such occurrence is called an inversion. Any permutation is called
even or odd according as the number of inversions occurring in it is even
or odd. The principal permutation, which has no inversion, is classed as
even. To count the number of inversions in a given permutation, take
76 PLANE ANALYTIC GEOMETRY [V, § 65
each subscript in order and see by how many higher subscripts it is
preceded. Thus, in the permutation
the subscript 1 is preceded by the higher subscripts 2, 3, 5 (3 inversions);
2 and 3 are preceded by no higher subscripts ; 4 is preceded by 5 (1 in-
version) ; 5, 6, 7 are not preceded by any higher subscripts. Hence there
are 3 -f 1 = 4 inversions, and the permutation is even. The permutation
of the same elements has 3 + 3 + 2 + 3 + 2 = 13 inversions, and is, there-
fore, odd.
66. If in a permutation any two adjacent elements are interchanged^
the number of inversions is changed by 1 ; hence the class to which the
permutation belongs is changed (from even to odd or from odd to even).
Let the two adjacent elements be ah, au and suppose that h<.k.
Two cases arise according as the original arrangement is ahak or ata/,.
(a) If the original arrangement is a^a^: i the new arrangement is a^rt/, ;
as A < k^ and as all other elements of the permutation remain unchanged,
the number of inversions is increased by 1.
(Z>) If the original arrangement is akCih , the new arrangement is anak
so that the number of inversions is diminished by 1."
67. If in a permutation any two elements lohatever be interchanged, the
number of inversions is changed by an odd number, and hence the class
of the permutation is changed.
Eor, the interchange of any two elements a^, ai, can be effected by a
number of successive interchanges of adjacent elements. If there are m
elements between an and ak, we have only to interchange a^ with the first
of these elements, then with the next, and so on, finally with ak, and
then ak with the last of the m elements, with the next to the last, and so
on ; thus in all wi + 1 + m = 2 wi + 1 interchanges of adjacent elements
are required, i.e. an odd number.
68. Of the n ! permutations of n elements just one half are even, the
other half are odd.
This follows by observing that if in each of the n ! permutations we
interchange any two elements, the same in all, every even permutation
V, §69] DETERMINANTS OF ANY ORDER
77
becomes odd and every odd permutation becomes even, and no two differ-
ent permutations are changed into the same permutation. After this
interchange we must have exactly the same n ! permutations as before.
Hence the number of even permutations must equal that of the odd
permutations.
The propositions about inversions are important for the theory of de-
terminants of the nth order to which we now proceed.
69. General Definition of Determinant. When n^ numbers are
given (e.g. the coefficients of the variables in n linear equations), arranged
in a square array, we denote by the symbol
«ii ••• «i,
and call determinant of the nth order the algebraic sum of the n ! terms
obtained as follows : the first term is the product of the n numbers in the
principal diagonal aiia22«33 ••• «„n ; the other terms are derived from this
term by permuting in all possible ways either the second subscripts or
the first subscripts, and multiplying each term by + 1 or — 1 according
as it is an even or odd permutation (i.e. contains an even or odd number
of inversions) .
It follows at once that every term contains n factors, viz. one and only
one from each row, and one and only one from each column.
It is readily seen that this definition gives in the case of determinants
of the second and third order the expressions previously used as defining
such determinants. For a determinant of the fourth order,
«11
an
ai3
au
an
«22
^28
^24
azi
az2
ass
a34
an
«42
^43
au
we obtain the 1 • 2 • 3 • 4 = 24 terms from the principal diagonal term
ana^asiau
by forming all the permutations, say of the second subscripts 1, 2, 3, 4
and assigning the + or — sign according to the number of inversions. If
these permutations are derived by successive interchanges of two sub-
scripts the terms will have alternately the + and — sign.
78 PLANE ANALYTIC GEOMETRY [V, § 70
70. The properties of the determinant of the nth order are essentially
the same as those of the determinant of the third order (§§ 44-47).
(1) The determinant is zero whenever all the elements of any row, or
all those of any column, are zero.
For, every term contains one element from each row and one from
each column.
(2) It follows from the same observation that if all elements of any
row {or of any column) have a factor in common, this factor can be taken
out and placed before the determinant.
(8) The value of a determinant is not changed by transposition; i.e.
by making the columns the rows, and vice versa, preserving their
order.
For, this merely interchanges the subscripts of every element, i.e. the
first series of subscripts becomes the second series, and vice versa.
Hence any property proved for rows is also true for columns.
(4) The interchange of any two rows (columns) reverses the sign of
the determinant.
For, the interchange of any two rows gives an odd number of inver-
sions to the first series of subscripts in the principal diagonal (§ 67), and
does not alter the second series. Hence the signs of all the terms are
reversed.
CoR. 1. A determinant in which the elements of any row {column)
are equal to the corresponding elements of any other row {column) is zero.
For, the sign of the determinant is reversed when any two rows
(columns) are interchanged ; but the interchange of two equal rows
(columns) cannot change the value of the determinant. Hence, denot-
ing this value by A, we have in this case — A = A, i.e. A = 0.
(5) If all the elements of any row (column) are sums of two terms, the
determinant can be resolved into a sum of two determinants.
For, in the expansion of the determinant every term contains one bi-
nomial factor ; therefore it can be resolved into two terms. See § 47 for
an illustration.
By means of this property, prove the following corollaries :
CoR. 1. If all the elements of any row (column) are algebraic
sums of any number of terms, the determinant can be resolved into a
corresponding number of determinants.
Cor. 2. The value of a determinant is not changed by adding to the
V, § 70] DETERMINANTS OF ANY ORDER
79
elements of any row (column) those of any other row {column) multiplied
by any common factor.
This corollary furnishes a method (see § 72) by which all the elements
but one of any row (column) can be reduced to zero.
EXERCISES
1. How many inversions are there in the following permutations ?
(a) aia^aza^aia^a^. (b) a7a6aiasa2aiaB. (c) a7a6«5«4a3a2«i-
2. In the expansion of the determinant below, what sign must be
placed before the terms celn, agjp ?
3. Show that
a
b
c
d
e
f
g
h
i
J
k
I
m
n
P
aix + biy + Ciz ai &i Ci
a^X + biy + C2Z «2 &2 C2
azx + bsy + Csz as 63 C3
a^x + biy + c^z a^ 64 C4
= 0.
4. Reduce the following determinant to
one in which all the elements
of the first column are
1:
2 4
1 3
3 7
5 6
2
5
6 1
2 3
5. Show that
(6 + c)2
a2
«2
(a)
62
(c + a)2
62
= 2 abc(a + 6 + c)^ ;
c2
c2 (a + 6)2
66' + cc'
ba'
ca'
(&)
ab'
cc' + aa'
c6'
= 4 aa'bb'cc'.
ac'
be'
aa
'+66'
(Hint. Multiply the rows by a, 6, c, respectively.)
80
PLANE ANALYTIC GEOMETRY
[V,§71
71. Minors and Cofactors. if in a determinant both the row
and column in which any particular element atu occurs be struck out, the
remaining elements form a determinant of order n — 1, which is called
the minor of the element aiu-
From the definition (§ 69), we observe that the expansion of any de-
terminant is linear and homogeneous in the elements of any one row
(column). The terms which contain an as factor are those terms whose
other elements have all possible permutations of either the first or second
subscripts 2, 3, ••• n. Hence the sum of the terms that contain an as
factor can be expressed as an multiplied by its minor, i.e.
«22 ••• din
«n • • •
dfii "' (Inn
By interchanging the first and second rows ((4) § 70) we observe similarly
that the sum of those terms which contain a-n as factor can be written
«i2 ••• ax,
— a'ix • • •
(ln2 '" Oni.
Those terms which contain a^i as factor are given by
«12 ••' «i«
asi . • .
a„2 ••• a,
and so on. Hence the expansion of a determinant by minors of the first
column is
«ii
a22 ••• a2r.
«n2
— «21
«12 ••• «ln
«n2
+ ... (- l)"+^a«i
«12
au
(hi-li 2 ""O^n-lj n
Let Aik denote the cofactor of atk ; that is, (— 1)»+* times the minor of Qik]
and A the original determinant ; we can then write this expansion in the
form
A — a\\A\\ 4- a>2iA2i + ^si^si + ••• + «ni^ni.
Similarly by cofactors of the elements of any column,
A = aikAik + a2*^2* + «3*^3fe + ••• + ankA„k, for A: = 1, 2, 3, ... n,
and by cofactors of the elements of any row,
A = anA i + a<2^i2 + aoAis -\- ... + ainAtn, for i = 1, 2, 3, ... n.
V, § 74] DETERMINANTS OF ANY ORDER 81
The evaluation of a determinant of order n is tlius reduced to the
evaluation of n determinants of order n — I. To each of these the same
process is applied until determinants of order 3 are obtained which can be
evaluated by the rule of § 42.
72. In case of numerical determinants this process of successive reduc-
tion is very much simplified by reducing to zero all the elements of any
one row (column) with the exception of one element, say Uik. This can
always be done by addition or subtraction of multiples of rows (columns),
by Cor. 2, § 70. The expansion by cofactors of the elements of this row
(column) then reduces to a single term, viz. aikAik.
The sign (— 1)«+* to be affixed to the minor of au to obtain the cof actor
Aik is readily found by counting plus, minus, plus, minus, etc., from the
first element an down to the itli row and then to the yfcth column until
Uik is reached.
73. The sum of the elements of any row (column) multiplied respec-
tively by the cofactors of the elements of any other row (column) is zero.
For, this corresponds to replacing the elements of any row (column) by
the elements of another row (column). Hence the determinant vanishes
(§ 70, (4), Cor. 1). For example, if in the expansion by cofactors of the
first row
aii^n + auAu + ••• + otiu^in
we replace the elements of the first row by those of any other row we find
anAn + ai2^i2 + ••• + «m^in = 0, for i = 2, 3, ••• w.
74. Linear Equations. We write n equations in n variables
aji, X2, Xs, ••• Xn as follows,
auXi + ttnXz + ••• -f ainXn = h,
anXl + a22X2 + ••• + a2nOf7i = ^2,
«nia^l + an2X2 + ••• + ann^n = kn-
The determinant formed by the coefficients of the variables is called the
determinant of the equations (§§37, 43) and is denoted by A. To solve
the equations for any one of the variables, say Xj, we multiply the first
equation by the cof actor of aij in A, i.e. by Ay, the second equation by
A2j, the third equation by Asj, etc., and add. This sum is by § 71
{aijAij + a2jA2j + ••• + anjAnj)Xj = Axj = kiAij + A;2^2j + ••• + knAnj,
as the coefficients of all the other variables vanish (§ 73). Hence if
82
PLANE ANALYTIC GEOMETRY
[V, § 74
^ :5£i 0, we have the following rule : Each variable is the quotient of two
determinants, the denominator in each case is the determinant of the
equations, while the numerator is obtained from the denominator by re-
placing the coefficients of the variable by the constant terms (§§ 37, 43).
75. Elimination, if the n linear equations are homogeneous, i.e. if
kiy ^2, '" kn are all zero, we have
«iia-'i + aiiX2 + ••• + ainXn = 0,
a^iXi + a22^2 + ••• 4- «2naJn = 0,
an\Xi + a„2aJ2 + ••• + ann^n = 0.
These equations are evidently satisfied by the values
Xi = 0, iC2 = 0, • • • iCw = 0.
Other values of the variables can satisfy the equations only if the deter-
minant of the equations is zero. For, the method of § 74 gives in the
case of homogeneous equations
Axi = 0, Axi = 0, ••• Axn = 0.
Hence if Xi, X2, ••• Xn are not all zero we must have
^ = 0.
This result may also be stated as follows : The result of eliminating n
variables between n homogeneous linear equations is the determinant of
the equations equated to zero.
If, for instance, Xn =^ 0, we can divide each equation by Xn and then
solve any n—1 equations for the quotients Xi/Xn-, Xz/Xn-, ••• Xn-i/Xn. It
thus appears as in § 48 that when ^ = the ratios of the variables can
be found unless all the cofactors Aij are zero.
EXERCISES
1. Show that
«ii
«12 «13 ai4
an an _
a2i
^22 «23 «24
an «22
1 au
1
2. Write the expansion of
X
as
-1 X
02
-1
X a\
— 1 ao
V, § 76] DETERMINANTS OF ANY ORDER
83
3. Express aox* + cli^^ + 0,2^'^ + «3aJ
4. Find the value of
054 as a determinant.
a
b
c d
— a
b
X y
— a
-b
c z
— a
-b
-c d
5. Show that
1 + a 1 1
1
1
1 1 + & 1
1
1
1 1 1 +
c 1
1
= abcde
1 11
1 + ^
1
111
1 1 + e
6. Solve the equations :
Sx+y- z
-2w=-S
,
(a) .
2x-y+5
5x + 4y-
z-Sw=6
z + w = 7,
>
(&)
. x+2y-
Sz +
w —-
3.
abcde (1+^ + 1 + ^ + 1 + 1)
\ a - b c d e J
ix-2y + 2z + w = ly
2x + Sy-Sz + Sw = 2,
X — y+z — 4:W=^y
Sx + y-4:z + Sw=-5.
7. Are the following equations satisfied by other values of the variables
than 0, 0, 0, ?
(a)
Sx-4:y + 5z+w = 0,
5x + 2y — Sz-io = 0,
X — y + z + w = 0,
2x + 2y-3z + Sw = 0.
(&)
[Sx + 2y + z-6w = 0,
9x + 9y + 6z-l0w=0,
2x + y - z + Sw = 0,
x + 2y + z + iw = 0.
8. The relations between the sides and cosines of the angles of a tri-
angle are a = 6 cos 7 4- c cos /3, & = c cos a + a cos 7, c = a cos /3 + 6 cos a ;
find the relation between the cosines of the angles.
76. Special Forms, in any determinant
an ••• ain
two elements are called cowjw^aie when one occupies the same row and
column that the other does column and row respectively ; thus the
element conjugate to anc is aui. The elements with equal subscripts an,
a22, ••• ann are called the leading elements; they are their own conju-
84
PLANE ANALYTIC GEOMETRY
[V, § 76
A determinant in which each element is equal to its conjugate (i.e.
ttik = aki) is called symmetric.
A determinant in which each element is equal and opposite in sign to
its conjugate (i.e. aik = — au) is called skew-symmetric \ the condition
implies that the leading elements are all zero.
A skeio-symmetric determinant of odd order is always equal to zero.
For, if we change the rows to columns (§70, Prop. 3) and multiply
each column by — 1, the determinant resumes its original form. But
since the determinant is of odd order we have multiplied by — 1 an odd
number of times, which changes the sign of the determinant [(4), § 70].
Hence denoting the value of the determinant by ^4, we have — A — A,
i.e. A=0.
77. Multiplication, it can easily be verified for determinants of
the second order that the product of any two such determinants
«ii .«i2| f>n hi
(221 ^22 I ^21 ^22
can be expressed as a determinant of the second order in any one of the
four following forms :
«ii?>ii + cinbu
(i2il>n + a22&i2
^ii^ii -f a2i&2i
ai2?>ii + a22&2i
dnbii +ai2&22
«21&21 + a22'^22
ail&12 + «21&22
ai2&12 4- ^22^22
anbii 4- aiibii dnbn + 012622
a21?>ll + dllbil a2lbi2 + a22&22
ail^U + «21&12
dnbn + diibii
anbii + 021^22 1
^12^21 + a22&22
Thus the first of these forms is, by (6), § 70, equal to the sum of four
determinants
«ii&ii
021611
011621! |aii6ii
021621! I021611
012622
^22622
012612
022612
011621
O21621
012612
O22612
012622 ]
022622
of which the first and fourth are zero, while the sum of the second and
third reduces to
611622
For determinants of higher order the same method can be shown to
hold. Without giving the general proof we here confine ourselves to
illustrating the' metho'd for determinants of the third ord^r :
On 012
a2i O22
— 612621
On 012
021 022
=
«ii
021
012
O22
611
621
612
622
V, § 77] DETERMUSTANTS OF ANY ORDER
85
an «i2 «i3
ftll &12 &13
Cii C12 Ci3
^21 «22 «23
&21 &22 &23
=
C21 C22 C23
«31 «32 «33
631 &32 &33
C31 C32 C33
where
Cn = «11&11 + «12&12 + «13&13» C12 = ail&21 + «12&22 + «13&23»
Cl3 = ail&31 + Clnhi + «13&33, C21 = a2lbn + «22&12 + «23^13i
etc. The product determinant can here also be written in four different
forms, according as we combine rows with rows, rows with columns,
columns with rows, or columns with columns.
If the two determinants to be multiplied are not of the same order,
they can be made of the same order by adding to the lower determinant
columns and rows consisting of zeros and a one ; thus
1
a h
a &
z=.
c d^
c d
etc.
EXERCISES
1. Show that (a) The minors of the leading elements of a symmetric
determinant are symmetric. (6) The minors of the leading elements of
a skew-symmetric determinant are skew-symmetric, (c) The square of
any determinant is a symmetric determinant.
2. Expand the symmetric determinants :
(«)
('0
H G
B F
F C
(&)
1
1
1
X
1
X
1
y
z
x 4- p px + q
x+p
px -^q
Show that
(a)
1
1
1
a
b
=
1
c
d
1 11
1 a-{- a & + /3
1 c+ a d + p\
(e)
(0
1 X
X 1
y
z
1
1
1
1
y z
(State this property in words.)
86
PLANE ANALYTIC GEOMETRY
[V, § 77
(?>)
X a
a a
X a a
a X a
= {x-ay\x-v2a).
(c)
a X
a a
a a
X a
a a X
a a
a X
= (a;-ffl)3(a;-f-3a).
4. Show that any determinant whose elements on either side of the
principal diagonal are all zero, is equal to the product of the leading
elements.
5. A symmetric determinant in which all the elements of the first
row and first column are 1 and such that every other element is the sum
of the element above and the element to the right of it, has the value 1.
Illustrate this proposition for a determinant of the fourth order.
6. Show that any skew-symmetric determinant of order 2 or 4 is a
perfect square. This is true for any skew-symmetric determinant of
even order,
7. Expand the following determinants :
a be
-a f e
-h -f (id
— c —e —d
(«)
1 a
-a 1
h -c
(&)
• (c)
a
-6
— a
/
h
-/
— c
— e
-d
8. Express as a determinant
(a)
id)
d\
(&)
(c)
X a a
1
a X a
.
1
a a X
1
(0
X
z
X
y
y
z
-11
si
(/)
an — S ai2 G5l3
«31 «32 Cf33
9. Show that
a b c
d e f
•
g h k
d'
b' c'
e' f
=
h' k'
aii4
s
«12
an
^21
«22 + S
«23
an
«32
a33 4-
a b
c
d e
f
g h
k
a p
7
a'
b'
c'
8 e
f
d'
e'
f
V e
K
g'
h'
k'
CHAPTER VI
y
(
h
I 1 X
—0
Fig. 31
THE CIRCLE. QUADRATIC EQUATIONS
78. Circles. A circle, in a given plane, is defined as the locus
of all those points of the plane which are
at the same distance from a fixed point.
Let C (h, k) be the center, r the radius
(Fig. 31) ; the necessary and sufficient
condition that any point P (x, y) is at
the distance r from C (h, k) is that
(1) (a? - hfj^{y _ Jc)^=r^,
This equation, which is satisfied by the coordinates x, y of
every point on the circle, and by the coordinates of no other
point, is called the equation of the circle of center C (h, k) and
radius r.
If the center of the circle is at the origin (0, 0), the equation
of the circle is evidently
(2) ar^+y = r2.
EXERCISES
Write down the equations of the following circles :
(a) center (3, 2), radius 7 ;
(6) center at origin, radius 3 ;
(c) center at (— a, 0), radius a ;
(d) circle of any radius touching the axis Ox at the origin ;
(e) circle of any radius touching the axis Oy at the origin.
Illustrate each case by a sketch.
87
88 PLANE ANALYTIC GEOMETRY [VI, § 79
79. Equation of Second Degree. Expanding the equation
(1) of § 78, we obtain the equation of the circle in the new form
x^ -\- y"" ~2hx-2ky -^-h} + l? - r^ = 0.
This is an equation of the second degree in x and y. But it is of
a particular form. The general equation of the second degree
in X and y is of the form
(3) Ax'^^ Ilxy + By^ + 2Ox-\-2Fy-\-C=^0;
i.e. it contains a constant term, (7; two terms of the first de-
gree, one in x and one in y ; and three terms of the second de-
gree, one in a^, one in xy, and one in y\
If in this general equation we have
it reduces, upon division by A, to the form
^ + f+^x + ^^y + ^ = 0,
which agrees with the form (1) of the equation of a circle, ex-
cept for the notation for the coefficients.
We can therefore say that any equation of the second degree
which contains no xy-term and in which the coefficients of a? and
y^ are equal, may represent a circle.
80. Determination of Center and Radius. To draw the
circle represented by the general equation
(4) Ax^ + Ay^ ^2Gx-\-2Fy^C = 0,
where A, G, F, C are any real numbers while ^ ^^ 0, we first
divide by A and complete the squares in x and y ; i.e. we first
write the equation in the form
, GW f , FY G\ F""
'^'-aJ^'^aJ-a^^a^'a
The left-hand member represents the square of the distance of
the point (x, y) from the point {—G/A, —F/A)\ the right-
VI, § 81] THE CIRCLE. QUADRATIC EQUATIONS 89
hand member is constant. The given equation therefore repre-
sents the circle whose center has the coordinates
h- ^ k- ^
and whose radius is
This radius is, however, imaginary \i G^ -{- F"^ <, AG ] in this
case the equation is not satisfied by any points with real co-
ordinates.
If G^ + F^ = AG, the radius is zero, and the equation is satis-
fied only by the coordinates of the point ( — G/A, — F/A).
If G^+F^ > AG, the radius is real, and the equation repre-
sents a real circle.
Thus, the general equation of the second degree (3), § 79, repre-
sents a circle if, and only if,
A = B^O,'H^O, G' + F'>AG.
81. Circle determined by Three Conditions. The equation
(1) of the circle contains three constants h, k, r. The general
equation (4) contains four constants of which, however, only
three are essential since we can always divide through by one of
these constants. Thus, dividing by A and putting 2 G jA = a,
2 F/A = b, C/A — c, the general equation (4) assumes the form
(5) ay'^.f^axi-by-^c^O,
with the three constants a, b, c.
The existence of three constants in the equation corresponds
to the possibility of determining a circle geometrically, in a
variety of ways, by three conditions. It should be remembered
in this connection that the equation of a straight line contains
two essential constants, the line being determined by two
geometrical conditions (§ 30).
90 PLANE ANALYTIC GEOMETRY [VI, § 81
EXERCISES
1. Draw the circles represented by the following equations:
(a) 2x^ + 2y^-Sx + 5y + l=0. (b) Sx^-\- Sy^+IT x - 16y-6 = 0.
(c) 4 ic2 + 4 2/2 _ 6 X - 10 y + 4 = 0. (d) x^ + y^ -\- x - 4:y =0.
(e) 2 x2 + 2 2/2 - 7 a; = 0. (f)x^-{-y^-Sx-6z=0.
2. What is the equation of the circle of center {h, k) that touches the
axis Ox ? that touches the axis Oy ? that passes through the origin ?
3. What is the equation of any circle whose center lies on the axis
Ox ? on the axis Oy? on the line y= x? on the line y = 2x? on the line
y = mx ?
4. Find the equation of the circle whose center is at the point (— 4, 6)
and which passes through the point (2, 0).
5. Find the circle that has the points (4, — 3) and ( — 2, — 1) as ends
of a diameter.
6. A swing moving in the vertical plane of the observer is 48 ft. away
and is suspended from a pole 27 ft. high. If the seat when at rest is 2 ft.
above the ground, what is the equation of the path (for the observer as
origin)? What is the distance of the seat from the observer when the
rope is inclined at 45^ to the vertical ?
7. Find the locus of a point whose distance from the point (a, h) is /c
times its distance from the origin.
Let P (ic, y) be any point of the locus ; then the condition is
V(x-a)2-f (2/-6)2= K Vx2 + 2/2 ;
upon squaring and rean*anging this becomes :
(1 - k2)x2 + (1 - k2)2/2 -2ax-2hy -\- cfi-\- 62 = o.
Hence for any value of k except k = 1, the locus is a circle whose center is
a/{\ - k2), 6/(1 - k2) and whose radius is k y/d^ + 6V(1 - k^). What
is the locus when ic = 1 ?
8. Find the locus of a point twice as far from the origin as from the
point(6, — 3). Sketch.
9. What is the locus of a point whose distances from two points Pi,
P2 are in the constant ratio k ?
VI, §82] THE CIRCLE. QUADRATIC EQUATIONS 91
10. Determine the locus of the points which are k times as far from
the point (—2, 0) as from the point (2, 0). Assign to k the values
\/5, V8, V2, I VS, ^ \/3, I \/2 and illustrate with sketches drawn with
respect to the same axes.
11. Determine the locus of a point whose distance from the line
Sx — 4y+l=0 is equal to the square of its distance from the origin.
Illustrate with a sketch.
12. Determine the locus of a point if the square of its distance from
the line x + y — a = is equal to the product of its distances from the
axes.
82. Circle in Polar Coordinates. Let us now express the
equation of a circle in polar coordinates. If (7(ri, <^i) is the
center of a circle of radius a (Fig. 32)
and P(rj <^) any point of the circle,
then by the cosine law of trigo-
nometry ^"'-^ T'^ .
r^ + ri^ — 2 riV cos (<^ — <^i) = a\ Fig. 32
This is the equation of the circle since, for given values of ?-i,
<^i, a, it is satisfied by the coordinates r, 4> of every point of
the circle, and by the coordinates of no other point.
Two special cases are important:
(1) If the origin^ be taken on the circumference and the
.polar axis along a diameter OA (Fig. 33),
the equation becomes
^2 _f_ a2 — 2 ar cos <f> = a^,
i.e. r = 2 a cos <^.
This equation has a simple geometrical
interpretation : the radius vector of any
point Pon the circle is the projection of the diameter OA =2 a
on the direction of the radius vector.
(2) If the origin be taken at the center of the circle, the
equation is r = a.
%
92 PLANE ANALYTIC GEOMETRY [VI, § 82
^ EXERCISES
1. Draw the following circles in polar coordinates :
'"^(a) r = 10 cos 0. '~~- (b) r = 2a cos (0 — ^ ir). (c) r — sin 0.
((?) r = 6. (e) r = 7 sin (0 — | tt) . {f)r- 17 cos 0.
2. Write the equation of the circle in polar coordinates :
\a) with center at (10, ^tt) and radius 5 ;
(6) with center at (6, \ tt) and touching the polar axis ;
(c) with center at (4, | tt) and passing through the origin ;
(d) with center at (3, tt) and passing through the point (4, \ tt) .
■^ 3. Change the equations of Ex, (1) and (2) to rectangular coordinates
with the origin at the pole and the axis Ox coincident with the polar axis.
4. Determine in polar coordinates the locus of the midpoints of the
chords drawn from a fixed point of a circle.
83. Quadratic Equations. The fundamental problem of
finding tlie intersections of a line and a circle leads, as we shall
see (§ 86), to a quadratic equation. Before discussing it we
here recall briefly the essential facts about quadratic equations.
The method for solving a quadratic equation consists in com-
pleting the square of the terms in x^ and a*, which is done most
conveniently after dividing the equation by the coefficient of x^.
The equation
«2 + 2 j9a; + g =
has the roots
x=. — J) ± V/52 — q.
The quantity i[P- — q\.^ called the discriminant of the equation.
According as the discriminant is positive, zero, or negative, the
roots are real and different, real and equal, or imaginary. In
the last case, i.e. when p^ < g, the roots are, more precisely,
conjugate complex, i.e. of the form a + bi and a — hi, where a
and h are real while i = V— 1.
As remarked above, any quadratic equation may be thrown
into the form here discussed, by dividing by the coefficient
of x^.
VI, § 84] THE CIRCLE. QUADRATIC EQUATIONS 93
84. Relations between Roots and Coefficients. If we de-
note the roots of the quadratic equation
by Xi and x^ , we have
Xi= — p -\- Vp^ — q, X2 = — p — Vi?2 — q,
whence
Xi -{- x^ = — 2 p, X1X2 = q ;
i. e. the sum of the roots of a quadratic equation in which the
coefficient ofx^ is reduced to 1 is equal to minus the coefficient of
x; the product of the roots is equal to the constant term.
With the values of x^, x.2 just given we find
{x — x^{;x — x^ = 0? -\- 2px + g,
so that the quadratic equation can be written in the form
{X — X^{X — CCg) = 0,
which gives
These properties of the roots often make it possible to solve
a quadratic equation by inspection.
EXERCISES
1. Solve the quadratic equations :
(a) a:2 - 6 X + 8 = 0. I (ft) x2 + 5 a; - 14 = 0.
(c) 2 a;2 - x - 28 = 0. ' (d) 6 jc^ - 7 a; - 6 = 0.
(e) a;2 + 2 &x - a^ + &2 ^^ q. {f) a^x^ - {a^ + b^)x + b'^ = 0.
(fir) ax2 + &x = 0. (h) 12 a:2 + 8 x - 15 = 0.
/ 2. Show that the solutions of the quadratic equation ax'^ + &x + c =
may be written in the form x = - -^ ± ^^ - 4 «c
2a 2a
When are these solutions real and unequal ? equal ? imaginary ?
3. Write down the quadratic equation that has the following roots :
(a) 3, - 2. (ft) - 3, 0. (c) 5, - 5.
(d) a-b, a + b. (e) 3 - 2V3, 3 + 2 >/3. (/) 1 + \/2, 1 - \/2^
(9) c, -i. (h) h-h ' (i) 3, V2.
94 PLANE ANALYTIC GEOMETRY [VI, § 84
4. Without solving, determine the nature of the roots of the follow-
ing equations :
» 5a;2-6x-2 = 0. (6) 9x^ + (>x+lz=0.
^c) 2 a:2 - a; + 3 = 0. (cZ) 20 a;2 + 6 a; - 5 = 0.
(e) llx2-4x-^^ = 0. (/) 3a:2 + 2x + l = 0.
6- For what values of k are the roots of the following equations real
and different ? real and equal ? conjugate complex ?
(a) x2- 4x + A: = 0. (6) a:2 + 2 ^•a; + 36 = 0.
i^ ,(c) 9x^ + kx+26 = 0. (d) ax^-\-bx + k = 0.
"^ (e) A:x2 - 5 X + 6 = 0. (/) ax2 + A:x + c = 0.
6. Solve the following equations as quadratic equations :
(_(a) ?/4_3y2_4^0. (Let 1/2= 2;.) (6) 2;3-2 + 3 2;-i - 2 = 0.
, , ,x 2 , X + 3 „
(c) X + V^TTS = 3. (^) ^^ + -^ = 2.
(e) m6 + 18 m3 - 243= 0. (/) 2 x"! + x'^ - 16 = 0.
7. If xi and X2 are the roots of x2 + 2 px + g = 0, find the values of
(7(a) Xi2x2 + X1X22. (6) Xi2 + X22. (C) (Xi - X2)2.
Xi X2 Xi2 X22
and apply these results to the case x2 — 3 x + 4 = 0.
8. Without solving, form the equation whose roots are each twice
the roots of x2 - 3 x + 7 = 0. [See § 84.]
9. What is the equation whose roots are m times the roots of
x2 + 2px + ^ = 0?
10. Form the equation whose roots are related to the roots of 2 x2 —
3 X — 5 = 0, in the following ways :
(o) less by 2 ; (h) greater by 3 ; (c) divided by 6.
85. Simultaneous Linear and Quadratic Equations. To
solve two equations in x and y of which one is of the first
degree (linear) while the other is of the second degree, it is
generally most convenient to solve the linear equation for either
X or y and to substitute the value so found in the equation of the
second degree. It then remains to solve a quadratic equation.
An equation of the first degree represents a straight line.
VI, §86] THE CIRCLE. QUADRATIC EQUATIONS 95
If the given equation of the second degree be of the form
described in § 79, it will represent a circle. By solving two
such simultaneous equations we find the coordinates of the
points that lie both on the line and on the circle, i.e. the points
of intersection of line and circle.
86. Intersection of Line and Circle. Let us find the in-
tersections of the line
y = mx -h b
with the circle about the origin
Substituting the value of y from the former equation into the
latter, we find the quadratic equation in x :
x^+(mx-{-by=r^,
or (1 + 'nv')^ + 2 mbx -\-b^-r^=:0'
The two roots Xi, X2 of this equation are the abscissas of the
points of intersection ; the corresponding ordinates are found
by substituting iCi, X2in y = mx + b.
It is easily seen that the abscissas Xi, x^ are real and differ-
ent if (l + mV-62>o,
. .0 b ^
I.e. II — ^:=z=: < r.
Vl + rn?
Since m = tan a, and hence 1/ Vl + m^ = cos a, the preceding
relation means that b cos a < r, i.e. the line has a distance from
the origin less than the radius of the circle. If
the roots x^, x^ are real and equal. The line and the circle then
have only a single point in common. Such a line is said to
touch the circle or to be a tangent to the circle. If
(1 + 'rri')7^ -b^<0,
the roots are complex, and the line has no points in common
with the circle.
96 PLANE ANALYTIC GEOMETRY [VI, § 87
87. The General Case. The intersections of the line and
circle
i»^ + 2/^ + «i» + &2/ + c = 0,
are found in the same way : substitute the value of y (or a;),
found from the equation of the line, in the equation of the
circle and solve the resulting quadratic equation.
It is often desired to determine merely ivhetlier the line is
tangent to the circle. To answer this question, substitute y
(or x) from the linear equation in the equation of the circle
and, without solving the quadratic equation^ write down the con-
dition for equal roots (p^ = q, § 83).
EXERCISES
1. Find the coordinates of the points where the circle x^ + y'^^ — x -\- y
— 12 = crosses the axes.
2. Find the intersections of the line 3aj + y— 5 = and the ^circle
x2 + 1/2 _ 22 a; - 4 y + 25 = 0.
3. Find the intersections of the line 2x — 1 y + 6 = and the circle
2x2 + 2y2 4.9x + 9?/-ll = 0.
4. Find the equations of the tangents to the circle xr + y'^ = 16 that
are parallel to the line y =—Sx -j-S.
5. Show that the equations of the tangents to the circle x^ -\- y"^ = r^
with slope m are y = mx ± rVl + m'^.
6. For what value of r will the line 3x-2y — 5 = 0be tangent to the
circle x^ + y^ = r^ ?
7. Find the equations of the tangents to the circle 2x'^ + 2y^ — Sx
+ 5?/ — 7 = that are perpendicular to the line x + 2y + 3 = 0.
8. Find the midpoint of the chord intercepted by the line 5x-y + 9=0
on the circle x^ -\-y^ = 18. (Use § 84.)
9. Find the equations of the tangents to the circle x^ + y2 _ 53 that
pass through the point (10, 4).
VI, §89] THE CIRCLE. QUADRATIC EQUATIONS 97
88. The Tangent to a Circle. The tangent to a circle (com-
pare § 86) at any point P may be defined as the perpendicular
through P to the radius passing through P. To find the equa-
tion of the tangent to a circle whose center is at the origin,
x^ -\- y"^ = r^,
at the point P (x, y) of the circle (Fig. 34), observe that the
distance p of the tangent from the origin
is equal to the radius r and that the
angle p made by this distance with the
axis Ox is such that
cos /? = - , sin /8 = -^ :
T r
substituting these values in the normal
form X cos /8 + r sin ^ = p of the Fig. m
equation of a line (§ 54), we find as equation of the tangent
xX-]-yY=r'^,
where x, y are the coordinates of the point of contact P and
X Y are those of any point of the tangent.
89. The General Case. To find the equation of the tangent
to a circle whose center is not at the origin let us write the
general equation (4), § 80, viz.
(4) Ax"" + .4?/2 + 2 (^a; + 2 i<V + C = 0,
in the form
F^ C
"+fT^i^+2j=^+^^ ^
a) a^
where — G/A, — F/A are the coordinates of the center and
Q2/ji + F^/A"- C/A is the square of the radius r (§ 80).
With respect to parallel axes through the center the same circle
has the equation
2.2 G^ , F""
-^ A'' A^
= r\
98 PLANE ANALYTIC GEOMETRY [VI, § 89
and the tangent at the point P{x, y) of the circle is (§ 88) :
Hence, transferring back to the original axes, we find as
equation of the tangent at P (x, y) to the circle (4) :
AxX-\-AyY-\-G{:x+X)^F{y^- Y)+ (7=0.
This general form of the tangent is readily remembered if we
observe that it can be derived from the equation (4) of the
circle by replacing x^ by xX, y^ hy yY, 2 a? by ic+ X, 2y'byy-\-Y.
EXERCISES \ \n
1. Find the tangent to the given circle at the given point :
(a) 0^2 + 2/2 = 41, (5, -4).
(6) x^ + y^ + Qx + ^y- 16 = 0, (-2, 3).
(c) 3a-2 + 3?/2 + 10a; + 17?/+18 = 0, (-2, -o).
(d) a;2 + ?/2 - ax -hy = 0, («, 6).
2. The equation of any circle through the origin can be written in the
form (§ 81) x^ + y^ + ax + by = 0; show that the line ax -\- by = is the
tangent at the origin, and find the equation of the parallel tangent.
3. Derive the equation of the tangent to the circle {x—h)'^+{y—k)^=:r^.
4. Show that the circles x'^ + y^ — 6x + 2y + 2 = and x^ + y'^ — 4y
+ 2 = touch at the point (1, 1).
5. Find the tangents to the circle x^ + y^ — 2x — 10y-{-9 = at the
extremities of the diameter through the point (— 1, 11/2).
6. The line 2aj + 2/ = 10 is tangent to the circle x^ + y'^ = 20 ; what is
the point of contact ?
7. What is the point of contact if Ax -{- By -h C = is tangent to the
circle x^ + y^ = r'^?
8. Show that x — y — l = is tangent to the circle aj^ + ?/2 + 4 x
— 10 ?/ — 3 = 0, and find the point of contact.
9. By § 86, the line y = mx + & has but one point in common with
the circle x^ + ?/2 = r^ if ( 1 + m'^)r^ = b^ ; show that in this case the radius
drawn to the common point is perpendicular to the line y = mx -\- b.
VI, § 90] THE CIRCLE. QUADRATIC EQUATIONS 99
90. Circle through Three Pomts. To fiyid the equation of
the circle passing through three points Piix^^y^, -^2(^2? 2/2)5
A (^3 J 2/3)? observe that the coordinates of these points satisfy
the equation of the circle (§ 81)
(6) .T2^2/' + ^a^ + % + c = 0;
hence we must have
(J)
^i + Vi + «^i + &2/1 + c = 0,
^2 + 2/2^ + «^2 + &2/2 +c = 0,
.^i + Vz^ + «% + &2/3+ c = 0.
From the last three equations we can find the values of a, 6,
and c ; these values must then be substituted in the first equa-
tion.
In general this is a long and tedious operation. What we
actually wish to do is to eliminate a, b, c between the four
equations above. The theory of determinants furnishes a very
simple means of eliminating four quantities between four
homogeneous linear equations (§ 75). Our equations are not
homogeneous in a, 6, c. But if we write the first two terms in
each equation with the factor 1 : (a?^ -f y^) . 1, (x-^ 4- y-^) • 1, etc.,
we have four equations which are linear and homogeneous in 1,
a, b, c ; hence the result of eliminating these four quantities is
the determinant of their coefficients equated to zero. Thus the
equation of tJie circle through three points is
=
Compare § 49, where the equation of the straight line through
two points is given in determinant form.
^ + y'
X y
1
aa' + 2/i'
Xi 2/1
1
3^2^ + 2/2'
^2 2/2
1
x^ty^'
Xs 2/3
1
100 PLANE ANALYTIC GEOMETRY [VI, § 90
EXERCISES
1. Find the equations of the circles that pass through the points :
^ (a) (2,3), (-1,2), (0,-3).
-^(6) (0,0), (1,-4), (5,0).
(c) (0, 0), (a, 0), (0, b).
• 2. Find the circles through the points (3, — 1), (— 1, —2) which
touch the axis Ox.
^ 3. Find the circle through the points (2, 1), (— 1, 3) with center on
the line 3x — y + 2-0.
4. Find the circle whose center is (3, — 2) and which touches the
line 3a: + 4y-12 = 0.
6. Find the circle through the origin that touches the line
4x-5y- 14 = Oat (6, 2).
6. Find the circle inscribed in the triangle determined by the lines
24x-7?/ + 3=0, 3x-4«/-9 = 0, 5x + 12y-50 = 0.
7.' Two circles are said to be orthogonal if their tangents at a point of
intersection are perpendicular ; the square of the distance between their
centers is then equal to the sum of the squares of their radii. If the
equations of two intersecting circles are
x^ -\-y^ + aix + biy + Ci =0, and x^ + y^ + a^x + &22/ + C2 = 0,
show that the circles are orthogonal when aia2 + 6162 = 2(ci + C2).
8. Find the circle that has its center at (—2, 1) and is orthogonal to
the circle x^ + y^-6x + S = 0.
9. Find the circle that has its center on the line i/ = 3 x + 4, passes
through the point (4, — 3), and is orthogonal to the circle
x^ + y^ + lSx + 5y + 2 =0.
91. Inversion. A circle of center O and radius a being given
(Fig. 35), we can find to every point P of the plane
(excepting the center O) one and only one point P'
on OP, produced beyond P if necessary, such that
OP . OP' = a2.
The point P' is said to be inverse to P with respect
to the circle (0, a) ; and as the relation is not Fig. 35
VI, §92] THE CIRCLE. QUADRATIC EQUATIONS 101
changed by interchanging P and P', the point P is inverse to P'. The
point is called the center of inversion.
It is clear that (1) the inverse of a point P within the circle is a point
P' without, and vice versa ; (2) the inverse of a point of the circle itself
coincides with it ; (3) as P approaches the center 0, its inverse P' moves
off to infinity, and vice versa.
The inverse of any geometrical figure (line, curve, area, etc.) is the
figure formed by the points inverse to all the points of the given figure.
92. Inverse of a Circle. Taking rectangular axes through O
(Fig. 36), we find for the relations between the coordinates of two in-
verse points P{x, y), P' (x', y'), if we put OP = r, OP' = r' ;
X y r
rr'
a2
r2
since rr' = a^ . hence
X'- ^'"^
y'--
X2+2/2'
and similarly
._ aV
11
_ a'^y'
X'2 + 2/'2
Fig. 36
These equations enable us to find to any curve whose equation is given the
equation of the inverse curve, by simply substituting for x, y their values.
Thus it can be shown that hy inversion any circle is transformed into
a circle or a straight line.
For, if in the general equation of the circle
^(x2 + y2) + 2 ^x + 2 Py + (7 =
we substitute for x and y the above values, we find
Aa^
x'2 + y'
+ 2(?a2.
4-2Pa2.
y'
4-c = o,
(X'2 + ?/'2)2 ■ x'-2 + y'-^ X'2 4- ?/'2
that is, Aa^ + 2 QaH' + 2 Fay + 0(x'2 + y''^) = 0,
which is again the equation of a circle, provided C ^0. In the special
case when C = 0, the given circle passes through the origin, and its in-
verse is a straight line. Thus every circle through the origin is trans-
formed hy inversion into a straight line. It is readily proved conversely
that every straight line is transformed into a circle passing through the
origin ; and in particular that every line through the origin is transformed
into itself, as is obvious otherwise.
102
PLANE ANALYTIC GEOMETRY
[VI, § 92
EXERCISES
1. Find the coordinates of the points inverse to (4, 3), (2, 0), (—5, 1)
with respect to the circle x^-j-y'^ = 26.
2. Show that by inversion every line (except a line through the center)
is transformed into a circle passing through the center of inversion.
3. Show that all circles with center at the center of inversion are
transformed by inversion into concentric circles.
4. Find the equation of the circle about the center of inversion which
is transformed into itself.
6. With respect to the circle x^ + y"^ = 16, find the equations of the
curves inverse to :
(a) x=b, (b) x-y=0, (c) x'^ + y^-6x=0, (d) x^+y^-lOy + l=0,
(e) Sx-^y-\-Q=0.
6. Show that the circle Ax'^ + Ay'^ -^2 Gx-\-2 Fy + a^A = is trans-
formed into itself by inversion with respect to the circle a:^ + y2 — q2^
7. Prove the statements at the end of § 92,
93. Pole and Polar. Let P, P' (Fig. 37) be inverse points with
respect to the circle (O, a) ; then the perpen-
dicular I to OP through P' is called the polar of
P, and P the pole of the line Z, with respect to
the circle.
Notice that (1) if (as in Fig. 37) P lies within
the circle, its polar I lies outside ; (2) if P lies
outside the circle, its polar intersects the circle
in two points ; (3) if P lies on the circle, its
polar is the tangent to the circle at P.
Fig. 37
Referring the circle to rectangular axes through its center (Fig. 38) so
that its equation is
x2 -)- 2/2 = a%
we can find the equation of the polar I of
any given point P(ic, y). For, using
as equation of the polar the normal
form X cos /3+ F sin /3 =;;, we have
evidently, if P' is the point inverse
toP:
VI, §94j THE CIRCLE. QUADRATIC EQUATIONS 103
cos/3
\/x^ + y'^
sin/3
therefore the equation becomes
xX
■vx'^ + y'^
yY _
p=OP' =
or simply
xX-\-yY=a'^.
This then is the equation of the polar I of the point P {x, y) with re-
spect to the circle of radius a about the origin. If, in particular, the
point P (a;, y) lies on the circle, the same equation represents the tan-
gent to the circle xP- ■\-'f — a^ at the point P (x^y), as shown previously
in § 88.
94. Chord of Contact. The polar l of any outside point P with
respect to a given circle passes through the points of contact Ci , C2 of
the tangents drawn from P to the circle.
To prove this we have only to show that if Ci is one of the points of
intersection of the polar I of P with the circle, then the angle OCiP
(Fig. 39) is a right angle. Now the triangles
OCiP and OP'Ci are similar since they have
the angle at in common and the including
sides proportional owing to the relation
OP • OP' = a2,
OP^ a
a 0P'\
i.e.
where a = OCi. It follows that ^ OC\P= j.^^ 3^
0P'Ci = |7r.
The rectilinear segment C1C2. is sometimes called the chord of contact
of the point P. We have therefore proved that the chord of contact of
any outside point P lies on the polar of P.
It follows that the equations of the tangents that can he drawn from
any outside point P to a given circle can be found by determining the
intersections Ci , Ci of the polar of P with the circle ; the tangents are
then obtained as the lines joining Ci , C2 to P.
104
PLANE ANALYTIC GEOMETRY [VI, § 95
95. The General Case. The equation of the polar of a point
P (x, y) with respect to any circle given in the general form (4),
§ 80, viz.,
(4) Ax^- + Ay^-\-2Gx + 2Fy + C = 0,
is found by the same method that was used in § 89 to generalize the
equation of the tangent. Thus, with respect to parallel axes through the
center the equation of the circle is
C
A'
the equation of the polar of P(x, y) with respect to these axes is by
--^-l-f
§93:
Hence, transferring back to the original axes, we find as equation of the
polar of P {x, y) with respect to the circle (4) :
AxX-\- AyY -{- G{x + X)+ F(y + Y)+ C = 0.
If, in particular, the point P (x, y) lies outside the circle, this polar
contains the chord of contact of P; if P lies on the circle, the polar be-
comes the tangent at P (§ 89).
96. Construction of Polars. if a point Pi describes a line I, its
polar h with respect to a given circle (0, a) turns about a fixed point,
viz., the pole P of the line I (Fig. 40).
Conversely, if a line h turns about one
of its points P, its pole Pi with respect
to a given circle {0, a) describes a line Z,
viz. the polar of the point P.
For, the line I is transformed by in-
version with respect to the circle (0, a)
into a circle passing through and
through the pole P of I; as this circle
must obviously be symmetric with respect
to OP it must have OP as diameter. Any
point Pi of I is transformed by inversion
into that point Q of the circle of diameter OP at which this circle is in-
tersected by OPi . The polar of Pi is the perpendicular through Q to
OPi ; it passes therefore through P, wherever Pi be taken on ^
The proof of the converse theorem is similar.
Fig. 40
VI, §96] THE CIRCLE. QUADRATIC EQUATIONS 105
The pole Pi of any line h can therefore be constructed as the intersec-
tion of the polars of any two points of h ; this is of advantage when the
line h does not meet the circle. And the polar h of any point Pi can be
constructed as the line joining the poles of any two lines through Pi ; this
is of advantage when the point Pi lies inside the circle.
EXERCISES
1. Find the equation of the polar of the given point with respect to
the given circle and sketch if possible :
(a) (4, 7),x2 + ?/2^8.
(6) (0, 0),x2 + ?/2-3x-4 = 0.
(c) (2, l),x2 + «/2_4x-2?/+l=0.
(rZ) (2, -3), x2 + 2/2+ 3a; +10?/+ 2 = 0.
2. Find the pole of the given line with respect to the given circle and
sketch if possible :
(a) X + 2 y - 20 = 0, a;2 + y/2 = 20.
(6) X + ?/ + 1 = 0, x2 + ?/2 = 4.
(c) 4 X - ?/ = 19, x2 + y2 = 25.
(d) Ax + By + C = 0, x2 + ?/2 = r2.
(e) 2/ = mx + 6, x2 + i/2 = r^.
3. Find the pole of the line joining the points (20, 0) and (0, 10),
with respect to the circle x^ + y^ = 25.
4. Find the tangent to the circle x2+«/2-10x+4 2/+9=0 at (7, - 6).
5. Find the intersection of the tangents to the circle 2 x2 + 2 y^— 15 x
+ y — 28 = at the points (3, 5) and (0, — 4) .
6. Find the tangents to the circle x2 + ]/2 — 6x — 10 2/ + 2 = that
pass through the point (3, — 3) .
7. Find the tangents to the circle x^ + y2 _ 3 x + y — 10 = that pass
through the point (— f, — V")-
8. Show that the distances of two points from the center of a circle
are proportional to the distances of each from the polar of the other.
9. Show analytically that if two points are given such that the polar
of one point passes through the second point, then the polar of the second
point passes through the first point.
10. Find the poles of the lines x - y -S = and x + y + S = with
respect to the circle x2 + 2/- _ 6 x + 4 y + 3 = 0.
106
PLANE ANALYTIC GEOMETRY [VI, § 97
If in the left-hand member of the equa-
97. Power of a Point.
tion of the circle
we substitute for x and y the coordinates xi , ?/i of a point Pi not on the
circle (Fig. 41), the expression (xi — hy -\- (yi — k)^ — r'^ is different
from zero. Its value is called the power y
of the point Pi (xi , y{) with respect to
the circle. As (x\ — h)'^ + {y\ — k)^ is
the square of the distance PiC = d be-
tween the point Pi {xi , yi) and the
center C(h, k), the power of the point
Pi (iCi, yi) with respect to the circle is
cP — r^; and this is positive for points
without the circle (d>r), zero for points Fig. 41
on the circle (d = r), and negative for points within the circle (d<ir).
If the point lies without the circle, its power has a simple interpretation ;
it is the square of the segment PiT = t of the tangent drawn from Pi to
the circle :
«2=(?2
(^i - hy -H (yi - ky - r2.
Hence the length t of the tangent that can be drawn from an outside
point Pi (a^i , yi) to a circle x'^ + y'^ -\- ax -\- hy -\- c = () i^ given by-
fa = xi2 + y{^ + axi + hyi + c.
Notice that the coefficients of x^ and y^ must be 1. Compare the similar
case of the distance of a point from a line (§ 56).
98. Radical Axis. The locus of a point whose powers with respect
to any two circles
x2 -}- 2/2 + axx + hiy + ci = 0,
a;2 + y2 + a^x + h^y + ca = 0,
are equal is given by the equation
a;2 + y2 + a^x + hiy + ci = x^-\-y'^ + a^x -f b^y + cs,
which reduces to
(ai — a2)x + (&i — h2)y + (t'l — ci) - 0.
This locus is therefore a straight line ; it is called the radical axis of the
two circles. It always exists unless ai = a<i and hi = ?)2, i-e- unless the
circles are concentric. - •
VI, § 99] THE CIRCLE. QUADRATIC EQUATIONS 107
Three circles taken in pairs have three radical axes which pass through
a common point, called the radical center. For, if the equation of the
third circle is
x2 + y2 + asx + hy + C3 = 0,
the equations of the radical axes will be
(a2 - as)x + (62 - b3)y + (C2 - C3) = 0,
(as - ai)x 4- (&3 - &i)y + (C3 - ci) = 0,
(ai - a2)x + (61 - b2)y + (ci - C2) = 0.
These lines intersect in a point, since the determinant of the coefficients
in these equations is equal to zero (Ex. 10, p. 57).
99. Family of Circles. The equation
(8) (a;2 4- 2/2 + a,x + b^ + c,) + k^x"" -\-y'-\- a,x + 6^ + Cg) =
represents a family, or pencil, of circles each of which passes
through the points of intersection of the circles
(9) i«2 + ^2^aiaj + &i2/+Ci = 0,
and
(10) a;2 + 2/2 + a^x + h^ + c^ = (),
if these circles intersect. For, the equation (8) written in the
form
(1 + k)x2 + (1 _^ ^^^y2 _,. (ct^ ^ ^a2)x + (61 + Kh^)y + Ci + KC2 =
represents a circle for every value of k except k = — 1, as the
coefficients of x^ and y"^ are equal and there is no xy-iQvm (§ 79).
Each one of the circles (8) passes through the common points
of the circles (9) and (10) if they have any, since the equation
(8) is satisfied by the coordinates of those points which satisfy
both (9) and (10). Compare § b^. The constant k is called the
parameter of the family.
In the special case when k — — 1, the equation i§ of the first
degree and hence represents a line, viz. the radical axis (§ 98)
of the two circles (9), (10). If the circles intersect, the radical
axis contains their common chord.
108 PLANE ANALYTIC GEOMETRY [VI, § 99
EXERCISES
1. Find the powers of the following points with respect to the circle
aj2 -\-y'2 — Sx—2y=0 and thus determine their positions relative to the
circle: (2,0), (0,0), (0, -4), (3,2).
2. What is the length of the tangent to the circle : (a) x^ -i- y'^ + ax
+ by-\-c = from the point (0, 0), (6) {x - 2)2 + (:« _ 3)2 - 1 = from
the point (4, 4) ?
3. By § 97, t^=:d-^ — r^=(d-{-r)(d-r); interpret this relation
geometrically.
4. Find the radical axis of the circles x^ -\- y^ + ax+ by + c = and
x^ + y^ + bx -\- ay + c = and the length of the common chord.
5. Find the radical center of the circles x^-\-y^ — Sx + 'iy — 7=0,
ic2 + ?/2 _ 16, 2(a;2 -I- ?/2) _f. 6 X + 1 = 0. Sketch the circles and their radi-
cal axes.
6. Find the circle that passes through the intersections of the circles
a;2 _f. ^2 _|_ 5 3j _ and x^ -\- y'^ + x — 2 y — 5 = 0, and (a) passes through
the point (—5, 6), (h) has its center on the line 4x — 2y — l5 = 0,
(c) has the radius 5.
7. Sketch the family of circles x^ + y^ - 6 y + k{x^ + ^/^ + 3 ?/) = 0.
8. What family of circles does the equation Ax -{■ By + O + k(x^
+ y^ -\- ax -{■ by -\- c) = represent ?
9. Find the family of curves inverse to the family of lines y = mx + 6;
(a) with m constant and b variable, (b) with m variable and b constant.
Draw sketches for each case.
10. Show that a circle can be drawn orthogonal to three circles, pro-
vided their centers are not in a straight line.
11. Find the locus of a point whose power with respect to the circle
2 .^2 -f 2 ?/2 — 5 X + 11 y — 6 = is equal to the square of its distance from
the origin. Sketch.
12. Show that the locus of a point for which the sum of the squares of
its distances from the four sides of a square is constant, is a circle. For
what value of the constant is the circle real ? For what value is it the
inscribed circle ?
VI, § 99] THE CIRCLE. QUADRATIC EQUATIONS 109
13. Find the locus of a point if the sum of the squares of its distances
from the sides of an equilateral triangle of side 2 a is constant.
14. Show that the circle through the points (2, 4), (— 1, 2), (3, 0) is
orthogonal to the circle which is the locus of a point the ratio of whose
distances from the points (2, 4) and (— 1, 2) is 3. Sketch.
15. Show that the circles through two fixed points, say (-a, 0),
(a, 0), form a family like that of Ex. 8.
16. The locus of a point whose distances from the fixed points (—a, 0),
(a, 0) are in the constant ratio k (:^ 1) is the circle
x2 + 2/2 4- 2^-±-^ax + a- = 0.
1 — k2
Compare Ex. 9, p. 90. Show that, whatever k(:^ 1), this circle inter-
sects every circle of the family of Ex. 15 at right angles.
Parameters, in problems on loci it is often convenient to express
the coordinates x, y of the point describing the locus in terms of a third
variable and then to eliminate this variable. Thus, for any point on a
circle of radius a about the origin we have evidently
(a) X = a cos 0, y = a sin <f> ;
eliminating <p by squaring and adding we find
^•2 ^ y2 - ^2.
The variable <p is called the parameter; the equations (a) are the
parameter equations of a circle about the origin.
17. The ends ^, jB of a straight rod of length 2 a move along two per-
pendicular lines ; find the locus of the midpoint of AB.
18. One end vl of a straight rod of length a describes a circle of radius a
and center O, while the other end B moves along a line through 0. Taking
this line as axis Ox and as origin, find the locus of the intersection of
OA (produced) with the perpendicular to the axis Ox through B.
19. Four rods are jointed so as to form a parallelogram ; if one side is
fixed, find the path described by any point rigidly connected with the op-
posite side.
20. An inversor is any mechanism for describing the inverse of a given
curve. Peaucellier's cell consists of a linked rhombus APBP' attached
by means of two equal links OA, OB to a fixed point 0. Show that this
linkage is an inversor, with O as center.
CHAPTER VII
COMPLEX NUMBERS
PART I. THE VARIOUS KINDS OF NUMBERS
100. Introduction. The process of finding the points of in-
tersection of a line and a circle (§ 86) involves the solution of
a quadratic equation. The solution of such a quadratic equa-
tion may involve the square root of a negative number. Thus
the roots of ic^ — 2a; + 3 = are x = l ±^—2.
The square root, or in fact any even root, of a negative num-
ber is called 201 imaginary number; and an expression of the
form a + V— 6 in which a is any real number and b any posi-
tive real number is called a complex number.
We shall first recall briefly the successive steps by which, in
elementary algebra, we are led from the positive integers to
other kinds of numbers.
101. Fundamental Laws of Algebra. The so-called natural
numbers, or positive integers 1, 2, 3, 4, • • • form a class of
things for which the operations of addition and midtiplication
have a clear and well-known meaning. These operations are
governed by the following laws :
(a) the commutative law for addition and for multiplication :
a-\-b = b i- a, ab = ba\
(p) the associative law for addition and for multiplication :
(a + 6) -f c = a + (6 -f c), {ab)c = a{bc) ;
(c) the distributive law, connecting addition and multiplication :
{a -\- b) c = ac -\- be, a(b -\-c) = ab-\- ac.
110
VII, §103] COMPLEX NUMBERS 111
102. Inverse Operations. The result obtained by adding
or multiplying any two or more positive integers is alwaj^s
again a positive integer.
This is not true for the so-called inverse operations : subtrao-
tion, the inverse of addition, and division, the inverse of multi-
plication. To make these inverse operations always possible
the domain of positive integers is extended by introducing :
(a) the negative numbers and the number zero ;
(6) the (positive and negative) rational fractions.
The relation between these various kinds of numbers is best
understood by imagining -j i -si -i\ \o \i [g p ^
the positive integers repre- Fig. 42
sented by equidistant points on a line, or rather by the distances
of these points from a common origin O (Fig. 42).
Negative numbers are then represented by equidistant points
on the opposite side of the origin; zero is represented by the
origin ; and fractions correspond to intermediate points.
103. Rational Numbers. The positive and negative inte-
gers, the rational fractions, and zero, form the domain of ra-
tional numbers. By adopting the well-known rules of signs the
operations of addition and multiplication and their inverses,
subtraction and division, can be extended to these rational num-
bers ; and all four of these operations, with the single exception
of division by zero, can be shown to be always possible in the
domain of rational numbers, so that any finite number of such
operations performed with a finite number of rational numbers
produces again a rational number.
In the domain of positive integers such linear equations as
a;-f-7 = 0, 5a; — 3 = cannot be solved. But in the domain of
rational numbers the linear equation ax-\-b = can always be
solved if a and b are rational and a is not zero.
112 PLANE ANALYTIC GEOMETRY [VII, § 104
104. Laws of Exponents. In the domain of positive inte-
gers, we pass from addition to multiplication by denoting a
sum of h terms each equal to a by the symbol ah, called the
product of a and h. Similarly, we may denote a product of
b factors each equal to a by the symbol aJ* ; this operation is
called raising a to the bth poicer, or involution. By this defini-
tion, the symbol a^ has a meaning only when the exponent h is
a positive integer. But the base a may evidently be any
rational number. The laws of exponents, or of indices,
a^ ' a'' = ap+'^, a^ • 5^= (a^)p, {a^y = a^",
follow directly from the definition of the symbol a*. The re-
sult of raising any rational number to a positive integral power
is always a rational number.
105. The Inverses of Involution. It should be observed
that the symbol a^ differs from the symbols a + h and ah in
not being commutative (§ 101) ; i.e. in general a and h cannot
be interchanged:
a^ =/= 6", if h^ a.'
It follows from this fact that while addition and multiplication
have each but one inverse operation, involution has two :
(a) If in the relation
a^ = G
h and c are regarded as known, the operation of finding a is
called extracting the hth root of c, or evolution, and is expressed
in the form
a = Vc.
(h) If in the same relation a and c are regarded as known,
the operation of finding h is called taking the logarithm of c to
the hase a and is indicated by
b = log„ c.
Logarithms will be discussed in Chapter XII ; for the present
we shall consider only the former inverse operation.
VII, § 107] COMPLEX NUMBERS 113
106. Irrational Numbers. Even when a, h, and therefore c
are positive integers, the extraction of roots is often impossible,
not only in the domain of positive integers, bat even in the
domain of rational numbers. Thus, in so simple a case as
6 = 2, c = 2, we find that a = V2 is not a rational number, i e.
it is not the quotient of any two integers, however large. For,
suppose that V2 = h/k, where h and k are integers and the ra-
tional fraction h/k is reduced to its lowest terms ; then squaring
both sides, we find 2 = h^/k^. But the rational fraction h^/k"^
is also reduced to its lowest terms and consequently cannot
be equal to the integer 2.
We are thus led to a new extension of the number system
by including the results of evolution : any root of a rational
number that is not a rational number is called an irrational
number. The rational and irrational numbers together form
the domain of real numbers.
If numbers are represented by points on a line as in § 102,
the number V2 has a single definite point corresponding to it
on the line ; for, the segment representing it can be found as
the hypotenuse of a right triangle whose sides have the length 1.
It can be shown that a single definite point corresponds to
any given irrational number.
It thus appears that although the rational numbers, " crowd
the line," i.e. although between any two rational numbers, how-
ever close, we can insert other rational numbers, they do not
" fill " the line ; i.e. there are points on the line that cannot
be represented exactly by rational numbers.
107. Extension of Laws. A rigorous definition and dis-
cussion of irrational numbers requires somewhat long and com-
plicated developments. It will here suffice to state the result
that irrational numbers are subject to the same rules of operation
as are rational numbers.
114 PLANE ANALYTIC GEOMETRY [VII, § 107
The fundamental laws of addition and multiplication (§ 101)
hold therefore for all real numbers, and so do the laws of signs
of elementary algebra. As regards the laws of exponents
(§104), they can be shown to hold when the bases are any real
numbers. Moreover, it can be shown that the symbol a^ has
a definite meaning even when the exponent h is any real num-
ber, and that the laws of exponents hold for such powers, pro-
vided only that the bases are positive. It is known from
elementary algebra how this can be done for rational exponents
by defining the symbols a° and a~"* as
a« = l, a-^ = —:
a*"
and it is shown in the theory of irrational numbers that the
latter definition can be used even when m is irrational.
Thus the laws of exponents (§ 104) hold for any real ex-
ponents provided the bases are positive.
108. Measurement. Historically, the gradual introduction
of rational fractions, of negative numbers, of irrational num-
bers, was determined very largely by the ajyplications of arith-
metic and algebra. Any magnitude that can be subdivided
indefinitely into parts of the same kind as the whole, and
hence can be "measured," leads naturally to the idea of the
fraction. Magnitudes that can be measured in two opposite
senses, like the distance along a line, the height of the ther-
mometer above and below the zero point, credit and debit, the
height of the water level above or below a fixed point, suggest
the idea of negative numbers. The incommensurable magnitudes
that occur frequently in geometry lead to the introduction of
irrational numbers. One of the principal advantages of algebra
consists in the remarkable fact that all these different kinds of
numbers are subject to the same simple laws of operation.
VII, § 110] COMPLEX NUMBERS 115
109. Imaginary Numbers. As mentioned in § 107, there is
still a restriction, in the domain of real (i.e. rational and
irrational) numbers, to the use of the laws of exponents (§ 104) :
the square root of a negative number has no meaning in this
domain.
Thus, V— 2 is not a real number ; for, by the definition of
the square root, the square of V— 2 is — 2 ; but there exists
no real number whose square is — 2. In other words, such
simple equations as x^ + 2 = 0, a;^ — 2 a; + 3 = have no real
solutions. It has therefore been found of advantage to give one
further extension to the meaning of the term " number," by
including the even roots of negative numbers, under the name
of imaginary numbers.
110. The Imaginary Unit. Any even root of a negative
(rational or irrational) number is defined as an imaginary
number. Every such number can be reduced to the form
± V— a, where a is positive. It is customary to denote V — 1
by the letter / and call it the imaginary unit. Any imaginary
number ± V— a can therefore be written in the form
± V — a = ± Va i ;
that is, every imaginary number is a real multiple of the imag-
inary unit I. Notice that as i = V— 1 we always have
1-2 = - 1.
The algebraic sum of a real number and an imaginary num-
ber, i.e. the expression a + bi where a and b are real, is called
a complex number. Notice that the domain of complex num-
bers includes both real and imaginary numbers. For, the
complex number a + bi is real in the particular case when
6 = 0, it is an imaginary number if a = 0. The great advan-
tage of complex numbers lies in the fact that all the seven
fundamental operations of. algebra (viz. addition, subtraction,
116 PLANE ANALYTIC GEOMETRY [VII, § 110
multiplication, division, involution, evolution, and logarithmi-
zation), with the single exception of division by zero, can be
performed on complex numbers, the result being always a
complex number ; i.e. if we denote by a, (3 any two complex
numbers, then a -\- 13, a — (3, a(3, a/^, a^, -v/cc, log^ a can all be
expressed in the form a + bi. It can then be shown that every
algebraic equation of the nth degree has 7i complex roots.
111. Imaginary Values in Analytic Geometry, in elemen-
tary analytic geometry we are concerned with "real" points and lines,
i.e. with points whose coordinates are real and with lines whose equations
have real coefficients. But it should be observed that points with com-
plex coordinates may lie on real lines and that lines with complex coeflQ-
cients may contain real points. Thus, the coordinates of the point
(2 + 3 i, 1 — 2i) satisfy the equation of the real line 2x + Sy— 7 = 0,
and the equation (I + 2 i)x — {2 -^ S i) y -{• 1 = is satisfied by the point
(3, 2). Calculations with imaginary points and lines may therefore lead
to results about real points and lines.
A rather striking example is afforded by the the theory of poles and
polars with respect to the circle. We have seen (§§ 93-95) that with
respect to a given circle every line of the plane (excepting those through
the center) has a real pole and every point (excepting the center) has a
real polar. If the line I intersects the circle in two points §i , ^2 » its
pole P can be found as the intersection of the tangents at Qi, Q2. If the
line I does not intersect the circle, this geometrical construction is im-
possible. But the analytic process of finding the points of intersection of
the line I with the circle can be carried through. The coordinates of the
points of intersection will be imaginary ; and hence the equations of the
tangents at these points will have imaginary coefficients. But the point
of intersection of these imaginary lines will be a real point ; viz. the pole
P of the line I and its real coordinates can be found in this way.
Thus to find the pole of the line y = 2 with respect to the circle
x^-^y^ = l we obtain the imaginary points of intersection (VSi, 2) and
(— V3i, 2) ; the imaginary tangents at these points are therefore:
VSix -f- 2 y = 1, — VS ix + 2y = 1; these imaginary lines intersect in the
real point (0, ^); it is easy to show that this is the required pole.
Vn, § 113] COMPLEX NUMBERS 117
PART II. GEOMETRIC INTERPRETATION OF
COMPLEX NUMBERS
112. Representation of Imaginaries. The meaning of com-
plex numbers will best be understood from their graphical
representation.
We have seen (§ 102) that every real number a can be repre-
sented by a point ^ on a straight line on which an origin
and a positive sense have been selected. _ .
We shall call this line (Fig. 43) the
axis of real numbers, or briefly the real
axis. I I o \ BealAxi s
To represent the imaginary numbers
we draw an axis through O at right
angles to the real axis and call it the
axis of imaginary numbers, or briefly ^^' '
the imaginary axis. The point A' on this axis, at the distance
OA' — a from the origin, can then be taken as representing
the imaginary number ai.
113. Representation by Rotation. This representation is
also suggested by the fundamental rule for dealing with im-
aginary numbers that i* = — 1. For, if a be any real number
and A its representative point on the real axis, the real num-
ber — a has its representative point A' situated symmetrically
to A with respect to on the real axis ; in other words, the
segment OA' which represents — a can be regarded as ob-
tained from the segment OA that rejjresents a by turning OA
through two right angles about 0. Thus the factor — 1 = t^
applied to the number a, or rather to the segment OA, turns
it about through two right angles. This suggests the idea
that the factor V— 1 = ?*, applied to a, may be interpreted as
118 PLANE ANALYTIC GEOMETRY [VII, § 113
turning the segment OA through one right angle in the counter-
clockwise sense so as to make it take the position OA'. Indeed,
if the factor i be now applied to ai, i.e. to the segment OA', it
will turn OA' into OA" and produce ai^ = — a.
Turning OA" counterclockwise through a right angle, we
obtain the point A'" on the imaginary axis which represents
ai^ = — ai; and finally, turning OA"' counterclockwise through
a right angle we regain the starting point A which represents
ai^ = a.
114. Representation of Complex Numbers. A complex
number, i.e. an expression of the form
z=x + yi,
where x, y are real numbers while i is the imaginary unit
V— 1, is fully determined by the two real numbers x and y,
provided we know which of these is to be the real part. If
we take the real axis as axis Ox, the imaginary axis as axis
Oy, of a rectangular coordinate system
(Fig. 44), the numbers x, y determine
a definite point of the plane, and only
one. This point P{x, y) can therefore
be taken as representative of the com- ^ <? ^
plex number z—x-\- yi. ^ict. 44
This representation also agrees with the idea (§ 113) that the
factor i turns through a right angle. For if we lay off on the
real axis, or axis Ox, OQ = x, and on the same axis QR = y
we obtain OR = OQ + QR = x-\-y; and if we turn QR about
Q through a right angle into QP we obtain x + yi and reach
the point P.
To every complex number z = x -\- yi thus corresiDonds one and
only one point P(x, tj) ; to every point P(x, y) of the plane cor-
responds one and only one complex number z = x -{■ yi.
VII, §116] COMPLEX NUMBERS 119
The real numbers, and only these, have their representative
points on the axis Ox-^ the imaginary numbers have theirs on
the axis Oy. The origin (0, 0) represents the complex num-
ber 0-{- iO = 0.
115. Correspondence of Complex Numbers to Vectors. It
should be recalled that strictly speaking (§ 102) a real number x
is represented, not by a point A of the real axis, but by the
segment OA = x. Similarly the complex number z=:x-{-yi is
represented, strictly speaking, not by the point P (Fig. 44), but
rather by the radius vector OP, taken with a definite direction
and sense. Thus the complex number z = x-\-yi represents a
vector (see §§ 19-20), whose rectangular components are x
and y. It will be shown below that the addition and subtrac-
tion of complex numbers follow exactly the laws of the com-
position of (concurrent) forces, velocities, translations, etc., in
the same plane.
116. Equality of Complex Numbers. Two complex num-
bers Z]^ = x^-\- y^i and Z2 = x^-{- y^i are called equal, if, and only
if, their representative points coincide, i.e. z^ = z^ if
x^ = X2 and yi = y^,
just as two forces are equal only when their rectangular com-
ponents are equal respectively.
If we apply the ordinary rules of algebra to the equation
^i + 2/i^* = ^2 4- yii
we obtain
Xi-X2 = (2/2 - yi)i-
Now the real number x^ — x^ cannot be equal to the imaginary
number {y^ — y^i unless ajj — x^^O and 2/2 — 2/i = ; whence
again we find a^i = x^, y^ = y^.
It follows in particular that the complex number z = x -{- yi
is zero if, and only if, a; = and y = 0.
120
PLANE ANALYTIC GEOMETRY [VII, § 116
EXERCISES
1. Locate the points which represent the following complex numbers :
(a) 4-3 i. (6) 2 i. (c) - 1 - i. (d) 4.
(e) A + .li. if) f-li. {g) -10-ti. {h) -ii.
2. Find the values of m and n in the following equations :
(a) {m - n) + (to + w - 2)z= 0. (6) (m2+w2-25) + (w-w-l)i=0.
(c) w + ni = 3 — 2 1. ((?) mm = m^ - w^ -j- 4 i.
3. Show that
(a) l3 =_i, (6) 1-5 3= i9^ (c) I'C + i^ = 0, (c?) 1-4 - l6 = 2.
4. Show that the following relations are true, n being any positive
integer :
(a) i^' = \. (6) i^+^=-i. (c) i*«-?>+2 = 2.
5. Show that
(«) K— 1 + VS i) is a cube root of 1,
{h) J (4- 1 — y/Zi) is a cube root of — 1.
117. Addition of Complex Numbers. The sum of tivo
complex numbers Zi = Xi-\- y^i and Z2 — X2 + y^i is defined as the
complex number z= (xi-{- x^) -\- (2/1+2/2)* j ii^ other words, if (Fig.
45) Pi is the point that represents Zi and P^ the point that rep-
resents Z2, then the point P that repre-
sents the sum z = Zi-^Z2 has for its ab-
scissa the sum of the abscissas of Pj
and P2 and for its ordinate the sum of
the ordinates of Pj and P.^. It appears
from the figure that this point P is the
fourth vertex of the parallelogram of
which the other three vertices are the origin and the points
P„P2.
118. Analogy to Parallelogram Law of Vectors. By com-
paring §§ 19, 20 it will be clear that the addition of two com-
y
P
R^--^
'/^\
«! 1 .
~0
""^Qz
<4 9
Fig.
45
VII, § 119]
COMPLEX NUMBERS
121
plex numbers consists in finding the resultant OP of their
representative vectors OP^, OPj.. The vectors may be thought
of as forces, velocities, translations, etc. In the case of trans-
lations this composition of two successive translations into a
single equivalent translation is particularly obvious.
While a real number «= OQ represents a translation along
the axis Ox*, an imaginary number yi a translation along the
axis O2/, a complex number z — x-\-yi can be interpreted as
representing a translation OP in any direction (Fig. 44). The
succession of two such translations % = a^ -|- y^ represented by
OPx (Eig. 45) and z.2, = x^-\- y^i represented by OP^ is equivalent
to the single translation z= (a?, -^x^ -f- (y^ H-//2)*' represented
by OP.
It follows that the addition of any number of complex
numbers (Fig. 46) whose
representative vectors are
OPi, OP2, OP3, OP4 can be
effected by forming the
Fia. 46
polygon 0PiP2'Ps'P; the
closing line OP is the rep-
resentative vector of the
sum ; precisely as in finding
the resultant of concurrent
forces (§ 20).
119. Subtraction. The difference of two complex numbers
2;^ = iCi -f- y^i and 22 = ^2 + 2/2** *^ ^6-
fined as the complex number z = (a^j
— ^2) H- (2/1 — 2/2)*'- Its representative
point P is found geometrically by
laying off from P^ (Fig. 47) a seg-
ment PjP equal and opposite to
OP2, i.e. equal and parallel to P2O.
122 PLANE ANALYTIC GEOMETRY [VII, § 120
120. Multiplication. The product of two complex yiumhers
Zi = Xi + 2/l^ and z^ = X2 + 2/2^' ^^ found by multiplying these two
expressions according to the ordinary rules of algebra and observ-
ing that 1*2 = — 1. We thus find :
z^z^ = (% + 2/iO(^'2 + ^20 = ^1^2 + ^m + ^22/1* + 2/l2/2*^
= {^v«2 - 2/12/2) + (a^i2/2 + ^iV^h
which is a complex number. A geometric construction will
be given in § 124.
121. Conjugate Imaginaries. Two complex numbers that
differ only in the sign of the imaginary part are called con-
jugate complex numbers. Thus, the conjugate of 5-|-2i is
5 — 2 i ; that of — 3 — ^ is — 3 + i, etc. The radii vectores rep-
resenting two conjugate numbers are situated symmetrically
with respect to the real axis.
Tlie product of two conjugate complex numbers is a real
number; for
{x + yi){x - yi) =:x'^+ y\
Notice that the roots of a quadratic equation are conjugate
complex numbers.
122. Division. To form the quotient of two complex num-
bers we may render the denominator real, by multiplying both
numerator and denominator by the conjugate of the denomi-
nator. Thus :
?i _ ^\ + y\i ^ (ag + .ViOfe — .VaO _ a?ia?2— aa?/2^' + ^'2y\i+ ViVi
2^2 3^2 4-2/2** (X2^-y.2i){X2 — y2i) X2+y2
= (^^&±M^\ 4- ( ^2V\ — ^y^ i
\ x,^+y,' J \xi + yi )'
Here also the result is a complex number. A geometric con-
struction is indicated in § 125.
yil, § 122] COMPLEX NUMBERS 123
EXERCISES
1. Simplify the following expressions and illustrate by geometric
construction :
(a) (3-60 + (4-20. (6) (4 - 3i)-(2 + i).
(c) (6+0 + (3-20-(0. (d) (2-30-(-l+0-(3 + 50.
(e) (4)-(30. (/) (0 + (3-2i)-(6).
2. Write the following products as complex numbers and locate the
corresponding points :
(a) (V5 + iV6)(\/6+iV5). (6) (3-zV8)(V3-fiV2).
(c) (vrTT-vn^)2. (d) (Va-V^^)3.
3. Show that
. s l + 2i l-2i ^ 3 (&) (X + 2/0^ - (a; - 2/0^ = 4a;?/i.
*^ M + i 1-1 ■ (c) (ix+yiy + <ix-yiy = 2(x'^ + f)-12x^y^.
4. Write the following quotients as complex numbers and locate the
corresponding points :
(a)
2 + 3i
4-i
(^)^-^^
(c) ^-3\
^ ^ 6 + 3i
(1 + 0(1 + 20(1 + 30
(0 ' .
(/) - ^ .
l + 4i
^ ^ -7+2i
'^'^^3-4i
(d)
5. Verify by geometric construction that the sum of two conjugate
complex numbers is a real number and that the difference is an imaginary
number.
6. Evaluate the following expressions for ^i = 3 + 4 i and Z2 = — 2 + 6i
and check by geometric construction :
(a.) 01-6. (&) 2^2 + 3. (c) 6-501. (d) Si-{-2zi.
(e) 2i-\zi. (/) 2-202. {g) 1(1-^1). {h) -Si-z^.
(0 01+2 02. U) 3 01 + 02. (fc) 01-2 02. (l) Zo-lZi.
(??i) 01 + 502— 4 i. (w) 02—^01 + 3. (o) 5 — 01 — 02. (p)02 — 6 — f0i.
7. Let Xi and ri represent the projections of a force Fi on the axes
of X and y, respectively, and X2 and F2 those of a second force F2. Show,
by the parallelogram law, that the projections on the axes of the result-
ant (or sum) of Fi and F2 are Xi + X2 and Yi + T2.
8. From Ex. 7, show that the correct results are obtained if Fi is
represented by Xi + Yii, F2 by X2 + ¥21, and their resultant by
Fi-{-F2= (Xi + FiO + (X2 + Tzi) = (Xi + X2) + ( Ti + ¥2)1.
124
PLANE ANALYTIC GEOMETRY [VII, § 123
123. Polar Representation. The use of the polar coordinates r,
of the representative point P{Xj y) leads to simple
interpretations of multiplication, division, involution,
and evolution.
The distance OP = r (Fig. 48) is called the modulus
or absolute value of the complex number ; the vec-
torial angle is sometimes called the argument, phase ^
or amplitude.
Fig. 48
Since
we can write
The right-hand member of this equation is the polar form of the complex
number z = x -\- yi.
x = r cos and y = r &m. 0,
z — X -\-yi = r(cos + i sin 0) .
124. Products in Polar Form. The product of two complex
numbers z\ = ri(cos 0i + i sin 0i) and 02 = »*2(cos 02 + i sin 02) is
0i2r2=?'i(cos0i+tsin0i)r2(cos02 + isin02)
=rir2[(cos0icos02 — sin0isin02) +i(sin 0i cos02+cos0i sin 02)]
=rir2[cos(0i + 02) + isin(0i + 02)].
This shows that the modulus of the product of two complex numbers is
the product of the moduli, the amplitude of the product is the sum of the
amplitudes, of the factors.
The point P that represents the product of the complex numbers repre-
sented by the points Pi and P2 (Fig. 49) can be constructed as follows :
Let Po be the point on the axis Ox at unit distance
from the origin and draw the triangle OPqPi ;
on OP2 construct the similar triangle OP2P. The
point P thus located is the required point. For,
by construction the angle P2OP=0i, hence the
angle PoOP=: 01 + 02- Moreover, as the triangles
OPoPi and OP2P are similar, their sides are pro-
portional, i.e.
1 : n = r2 : OP, whence OP = rir2.
Fig. 41)
125. Quotients in Polar Form. For the quotient of the two
complex numbers zi = ri(cos 0i + i sin 0i) and 02 = r2(cos 02 + i sin 02)
we find by making the denominator real :
VII, § 125]
COMPLEX NUMBERS
125
£i _ n (cos 01 + i sin 0i) _ ri(cos <pi + i sin 0i) (cos 02 — i sin 02)
«2 >'2(cos 02 + I sin 02) r2(cos 02 + i sin 02) (cos 02 — i sin 02)
_ri (cos 01 cos 02 4- sin 0i sin 02) + ^^(sin 0i cos </)2 — cos 0i sin ^2)
r2 cos2 02 + sin2 02
= ^ [cos (01 - 02) +1 sin (01 -02)].
r2
Hence the modulus of the quotient z — zi/z^ is the
quotient of the moduli, the amplitude is the differ-
ence of the amplitudes of Zi and z^. Evidently the
point P that represents the quotient z = Z1/Z2
Fig. 50) can be located by reversing the geometric
construction given in § 124; i.e. by constructing
on the unit segment OPq the triangle OPqP similar
to the triangle OP2P1.
EXERCISES
1. Write the following complex numbers in polar form :
(a) 2 + 2V3 i. (&) - 3 + 3 V3 i. (c) 6-6 i. (d) - 5 i.
(e) 7. (/) -8. (^) 5\/3-5i. (h) -10-lOi.
2. Write the following complex numbers in the form x + yi:
(a) 3(cos30° + isin30^).
(c) 10(cos I TT + i sin I tt) .
(e) V2(cosi7r + isin^Tr).
(g) 7(cosO -f isinO).
(i) 2 V3(cos I w + 1 sin I tt).
(k) ll(cos ^ TT + I sin I w).
(6) 5(cos I IT -\- i sin ^ -it) .
(d) 4(cos I TT + I sin I tt).
(/) V3(cos f TT + I sin I tt) .
(h) 5(cos7r + isin7r).
(j) 5 V2 (cos I TT + 1 sin I tt) .
(0 8(cos75° + isin75°).
3. Put the following complex numbers in polar form, perform the
indicated multiplication or division, and write the result in the form
X + yi' Check by algebra and illustrate by geometry.
(a) (2V3+2 0(3 4-3V3 0. (^) (1 + 0(2 + 20-
(c) (-2-20(5 + 50-
(e) (1+V30(1-V30
2VS-2i.
U)
bi
1+i
\-i
(h)
(k)
4i
(d) (_4 + 4V3 0(-3-3\/3 0-
(/) (-2)(-3 0.
-7
5+ 5i
1
- \/3 - i
(O
3 + 3V3 4
(0 —■
126 PLANE ANALYTIC GEOMETRY [VII, § 125
4. Show that the modulus of the product of the complex numbers
a + hi and c + di is y/(^a^ + b'^)(^c^ + d^).
6. Show by geometric construction that the product of two conjugate
complex numbers is a real number.
6. Show how to locate by geometric construction the point which
represents the reciprocal of a complex number.
7. Show that the point P that represents a complex number z and
the point P' that represents the conjugate of the reciprocal \/z are inverse
points with respect to the unit circle about the origin.
8. With respect to the unit circle about the origin, find the complex
numbers representing the points inverse to
(a) 3 + 4i. (6) 3+V^^. (c) - 5 + 3 i. (d) 1 - 6 i.
9. Show that the ratio of two complex numbers whose amplitudes
differ by ± | tt is an imaginary number.
10. Show that the ratio of two complex numbers whose amplitudes
are equal or differ by ± t is a real number.
126. De Moivre'S Theorem. The rule for multiplying two com-
plex numbers (§ 124) gives at once for the square of a complex number
z = r(cos <f> + i sin 0) :
z^ = [r(cos0 + isin0)]2 = r^(^cos2 + isin2 0).
Multiplying both members by ^ = r(cos + i sin 0) we find for the
cube :
z^ = [r(cos + 1 sin 0)]^ = r3(cos 3 + i sin 3 0) .
This suggests that we have generally for the ?ith power of 0, n being
any positive integer :
zn —\r (cos + 1 sin 0)]" = r'»(cos n + i sin n 0).
This is known as de Moivre' s formula.
To complete the formal proof we use mathematical induction (§ 62) .
Assuming the formula to hold for some particular value of w, it is at once
shown to hold for w+ 1, by multiplying both members by
z = »'(cos0 + isin0)
which gives
0«+i=[r(cos0 + I sin 0)]"+^ = r"+i[cos (w + l)0 + isin (w + l)0].
As the formula holds for w = 2, it holds for n = 3, and hence for w = 4, etc.,
i.e. for any positive integer.
VII, § 128] COMPLEX NUMBERS 127
127. Generalization of De Moivre's Theorem. De Moivre's
formula can be shown to hold for any real exponent n. That it holds for
a negative integer is seen as follows :
If in the formula for the quotient z = Zi/z2 (§ 126) we put ri = 1,
01 = 0, we find
— = — (cos 02 — i sin 02),
or dropping the subscript 2 :
- = - (cos <t> — i sin 0) ,
z r
If we raise this complex number to the nth power {n being a positive
integer) , which can be done by § 126, we find
(i)'
z-^ = — (cos«0 — I sin w0),
which proves de Moivre's formula for a negative integral exponent.
If in de Moivre's formula (§ 126) we put
a
ntjy = d, 1"^ = p, and hence = -, r = v^,
n
where y/p is the positive nth root of the real number p, we obtain
I \//)f cos- + /sin- I I =/)(cos^ + isin 0),
i.e. [/)(cos d + i sin 6)^= Vplcos - + i sin-V
\ n n)
This shows that de Moivre's formula holds when the exponent is of the
form \/n. The extension to the case when the exponent is any rational
fraction is then obvious.
128. Imaginary Roots. The last formula gives a means of finding
an will root of any real or complex number. To find all the roots of a
complex number z = p(cos 6 -\- i sin d) we must observe that as
cos d = cos (6 + 2 TTw), sin d = sin (^ + 2 irm),
where m is any integer, the number z can be written in the form
z = p[cos (^ -f 2 7rm) + / sin (^ + 2 Trm)],
so that by § 127 its roots are given by
128
PLANE ANALYTIC GEOMETRY [VII, § 128
cos ^±-2^+ I sin ^±1^
If in this expression we give to m successively all integral values, it takes
just n different values, viz. those for 7>i = 0, 1, 2, ••• , n — 1 ; therefore any
complex number z = p(cos 6 -\- i sin 6) lias n roots, viz. :
Vpfcos^ + isin^V ^pfcos^^t^+isini±^),
\ n 111 V '* *i '
.. Vp[i
+ (n-l)2
+ i sin
g + (H-l)27r '
These n roots all have the same modulus \/p, while the amplitudes differ
by 2 ir/n. Hence the points representing these n
roots lie on a circle of radius Vp about the origin
and divide this circle into n equal parts.
For example, the three cube roots of 8 i are found
as follows. In polar form
f
+ 8 I = 8(cos ^ TT + i sin I tt) ;
by de Moivre's formula (§ 127) we have
[8(cos ^ TT + z sin ^ tt)] 3
= 2[cos iZ_i_2= + i sin i^^Jl^jmJ,
= 2[cos (i TT + f irm) + i sin (^ tt + | Trm)] ;
w = gives the root :
w = 1 gives the root : 2(cos f tt + i sin f tt) = 2( — ^\/3+i \) = — VS + i ;
w = 2 gives the root : 2(cos | tt + i sin | tt) = 2(0 + i ( - 1)) = — 2 i.
If we put w = 3, we get the first root again, wi = 4 gives the second root,
and so on. Thus there are three distinct cube roots of 8 i, viz. VS + i,
\/3 + 1, — 2 i. These roots are represented by the points Pi, P2, Pa*
respectively (Fig. 51).
Fig. 61
VII, §129] COMPLEX .NUMBERS 129
129. Square Roots. The particular problem of finding the square
root of a complex number a + &i can also be solved by observing that the
problem requires us to find a complex number x + yi such that
a + 6i = (x + yiy.
Expanding the square and equating real and imaginary parts, we find for
the determination of x and y the two equations
x^-y^=a, 2 xy = h.
Eliminating y between these two equations, we obtain
a;2 - — =a ; that is, a;* - ax2 - i fo2 _ q •
whence Xi^ = |(a + Va^ + h-), x-^ = ^(a - Va^ + 62).
Since x is to be a real number and hence x^ must be positive, and as
a<A/a2+62 (unless 6 = 0, which ^ould mean that the given number a + bi
is real), we must take Xi2 and not X2^. Hence
x=± v|(a+Va2+62).
These values of x are zero only when b=0 and a < ; for then Va^ = — a.
In this particular case we find y = ±V— a, and hence
Va + bi =± V— a i.
In the general case, when & ^t 0, we find from the equation 2xy = b for
each of the two values of x one value of y.
' EXERCISES
1. Show how to locate the square of a complex number by geometric
construction. Locate the cube.
2. Show geometrically that 8i (Fig. 51) is the product of the numbers
represented by the points Pi, P2, P3.
3. For zi = l -\-2i, Z2—-2 — i show that z{^ — z^^ ={z\ + 2^2) (2^1— 02)
and illustrate geometrically.
4. For the same numbers verify and illustrate geometrically that
{Z\ — ZiY -Z^^2 ZxZi + 02^.
6. Show how to locate the points that represent the square roots of a
complex number.
6. Locate by geometric construction in two "ways the points which
represent [r(cos + i sin 0)]^.
K
130 PLANE ANALYTIC GEOMETRY [VII, § 129
7. Put the following complex numbers in polar form, perform the in-
dicated operations, and check by geometric construction :
(a) (1 + V3i)2. (b) (-1 + 0^. (c) (-V3-i)2.
id) ( V3 + i)^. (e) ( - 1)^. (/) ( -0^.
(g) Vl+V3i. (/i) -yZ-l-y/Si. (i) ^-2-\-2VSi.
U) ^/-'^-Su (A) v/-4-h4i. (Z) ^/6il.
(m) V'^Hol. (w) \/87. (o) V(-3i)».
8. Find the square roots of each of the following complex numbers
by using the method of § 129 :
(a) 7 + 24 i. (6) 4 1. (c) -2(8 + 151).
(d) - 16. (e) j\(5 - 12 i). (/) 4 a6 + 2(a2 - &2)i-.
(gf) _ 2[2 a& + (a2 - 62) ^-j. (^) _ 4 ^2^,2 4. 2(0* - 6*) i.
9. Find the three cube roots of unity and show that either complex
root is the square of the other, i.e. if one complex root of unity is denoted
by w, the other is w2. The three cube roots of unity then are 1, w, u^.
10. If 1 , w, w2 are the cube roots of unity (see Ex. 9) show that :
(a) 1 = w^ = w^ = w^", n being an integer.
(6) l + w + a>2 = 0.
(C) (1 + w2)4 = W.
(d) (a;i> + a;2g) (0,2^ 4. ^^g) (p^ g) ^ p8 + ^3.
(e) (1 _a; + w2)(l + w-a,2) ^4.
(/) (1 _ w + a;2) (1 - 0,2 4- 0,4) (1 _ 0,4 + w8) = _ 8 0,.
11. Prove de Moivre's formula for n any rational fraction, i.e. show
that, if p, g, w, are integers,
[r(cos + i sin 0)]^ = /.« fcos^^Jl^^ + I sin£^±l^l
L g g J
12. Show by geometric construction that the sum of the three cube
roots of any number is equal to zero ; that the sum of the four fourth
roots is zero.
13. Solve the following equations and locate the points which repre-
sent the roots :
(a)a:2-l=0. • (6) x^ + 1 = 0. (c) x* - 1 = 0. (d)x^-l=0.
(e) a:« - 1 = 0. ( f) x^ - 27 = 0. (g) x^ + 1=0. (/i) x* + 16 = 0.
(i) x5 + 32 = 0. (j) x2 + a2 = 0. (A;) x^ + ^3 = 0. (0 x^ - 1 = 0.
CHAPTER VIII
POLYNOMIALS. NUMERICAL EQUATIONS
PART I. QUADRATIC FUNCTION — PARABOLA
130. Linear Function. As mentioned in § 28, an expression
of the form mx + h, where m and h are given real numbers
(m=fzO) while ic may take any real value, is called a linear
function of x. We have seen that this function is represented
graphically by the ordinates of the straight line
y = mx 4- b ;
b is the value of y for x = 0, and m is the slope of the line, i.e.
the rate of change of the function y with respect to x.
131. Quadratic Function. Parabola. An expression of
the form aa^ -\-bx + c in which a ^ is called a quadratic func-
tion of X, and the curve
y = ax"^ -\-bx-{-Cj
whose ordinates represent the function, is called a parabola.
If the coefficients a, b, c are given numerically, any number
of points of this curve can be located by arbitrarily assigning
to the abscissa x any series of values and computing from the
equation the corresponding values of the ordinates. This
process is known as plotting the curve by points ; it is some-
what laborious; but a study of the nature of the quadratic
function will show that the determination of a few points is
sufficient to give a good idea of the curve.
131
132
PLANE ANALYTIC GEOMETRY [VIII, § 132
Fig. 52
132. The Form y = ax". Let us first take 6 = 0, c = ; the
resulting equation
(1) y = ax"^
represents a parabola which passes through the origin, since
the values 0, satisfy the equation. This x>ardbola is symmet-
ric ivith respect to the axis Oy ; for, the values of y correspond-
ing to any two equal and opposite values of x are equal. This
line oi symmetry is called the axis of the
parabola ; its intersection with the parab-
ola is called the vertex.
We may distinguish two cases accord-
ing as a > or a < ; if a = 0, the equa-
tion becomes 2/ = 0, which represents the
axis Ox.
(1) If a > 0, the curve lies above the axis Ox. For, no matter
what positive or negative value is assigned to x, y is positive.
Furthermore, as x is allowed to increase in absolute value, y
also increases indefinitely. Hence the parabola lies in the first
and second quadrants with its vertex at
the origin and opens upward, i.e. is con-
cave upward (Fig. 52).
(2) If a<0, we conclude, similarly,
that the parabola lies below the axis Ox,
in the third and fourth quadrants, with
its vertex at the origin and opens down-
ward, i.e. is concave downward (Fig. 53).
Draw the following parabolas:
y = x',y = ^x',y^-^o?,y^\x'.
133. The General Equation. The curve represented by the
more general equation
(2) y = ojx? + hx •{- c
differs from the parabola y — a^? only in position. To see this
Fig. 53
VIII, §134] POLYNOMIALS — THE PARABOLA
133
we use the process of completing the square in x\ i.e. we
write the equation in the equivalent form
y
I.e.
y-
If we put
7. + ^) = K''^^)-
4
^'
2 a 4 a
Fig. 54
the equation becomes
y — k = a(x — hy,
and it is clear (§ 13) that, with reference to parallel axes
OiXi, Oi2/i through the point Oi (Ji, k) the equation of the
curve is y-^ = ax^ (Fig. 54). The parabola (2) has therefore
the same shape as the parabola y = ax"^ ; but its vertex lies at
the point {h, k), and its axis is the line x = h. The curve
opens upward or downward according as a > or a < 0.
134. Nature of the Curve. To sketch the parabola (2)
roughly, it is often sufficient to find the vertex (by completing
the square in x, as in § 133), and the intersections with the axes.
The intercept on the axis Oy is obviously equal to c. The in-
tercepts on the axis Ox are found by solving the quadratic
equation
ax"^ -{- bx -j- c = 0.
We have thus an interesting interpretation of the roots of any
quadratic equation : the roots of ax^ -f- 6rc + c = are the
abscissas of the points at which the parabola (2) intersects
the axis Ox. The ordinate of the vertex of the parabola
is evidently the least or greatest value of the function
y = ax^ -\-hx-\-c according as a is greater or less than zero.
134 PLANE ANALYTIC GEOMETRY [VIII, § 134
EXERCISES
1. With respect to the same coordinate axes draw the curves y = ax^
for a=2^ f, 1, I, 0, — ^, — 1, — |, — 2. What happens to the parabola
y = ax^ as a changes ?
2. Determine in each of tlie following examples the value of a so that
the parabola y = ax^ will pass through the given point :
(a) (2,3). (6) (-4,1). (c) (-2, -2). (d) (3,-4).
3. A body thrown vertically upward in a vacuum with a velocity of v
feet per second will just reach a height of h feet such that h = ^^^ v^.
Draw the curve whose ordinates represent the height as a function of the
initial velocity.
(a) With what velocity must a ball be thrown vertically upward to rise
to a height of 100 ft. ?
(6) How high will a bullet rise if shot vertically upward with an ini-
tial velocity of 800 ft. per sec. , the resistance of the air being neglected ?
4. The period of a pendulum of length I {i.e. the time of a small
back and forth swing) is r= 2iry/l/g. Take g = S2 ft. /sec. and draw
the curve whose ordinates represent the length I of the pendulum as a
function of the period T.
(a) How long is a pendulum that beats seconds (i.e. of period 2 sec.) ?
(6) How long is a pendulum that makes one swing in two seconds ?
(c) Find the period of a pendulum of length one yard.
5. Draw the following parabolas and find their vertices and axes :
(a) y = lx^-x + 2. (h) y = -lx^ + x. (c) y = 5x^ + lbx + 3.
(d) y = 2-x-x^ (e) 2/ = a;2 - 9. (f)y = -9- x\
(^) y=3a;2_6« + 5. (/i) y = |a;2 + 2a; - 6. {i) x'^ - 2x -y = () .
6. What is the value of h if the parabola y = x^ -\- bx — 6 passes
through the point (1, 5) ? of c if the parabola y = x!:^ --Qx -\- c passes
through the same point ?
7. Suppose the parabola y = ax^ drawn ; how would you draw y =
a (x+2)2 ? y = a(x-7)2 ? y = ax2 + 2 ? y = a.r2 - 7 ? y = ax2+ 2x4-3?
8. What happens to the parabola y = ax'^ + hx + c as c changes ?
For example, take the parabola y = x2 — x + c, where c = — 3, — 2, — 1,
0, 1, 2, 3.
VIII, §134] POLYNOMIALS — THE PARABOLA 135
9. What happens to the parabola y = ax- + bx -\- c as a changes ?
For example, take y = ax'^ — x — 6, where « = 2, 1, |, 0, — ^, — 1, — 2.
10. (a) If the parabola y = ax^ + bx is to pass through the points
(1, 4), (— 2, 1) what must be the values of a and b ? (6) Determine the
parabola y = ax^ + bx + c so as to pass through the points (1, ^), (3, 2),
(4, f ) ; sketch.
11. The path of a projectile in a vacuum is a parabola with vertical
axis, opening downward. With the starting point of the projectile as
origin and the axis Ox horizontal, the equation of the path must be of the
form y = ax^ + bx. If the projectile is observed to pass through the points
(30, 20) and (50, 30), what is the equation of the path? What is the
highest point reached ? Where will the projectile reach the ground ?
12. Find the equations of the parabolas determined by the following
conditions :
(a) the axis coincides with Oy, the vertex is at the origin, and the
curve passes through the point (—2, — 3) ;
(6) the axis is the line x = 3, the vertex is at (3, — 2), and the curve
passes through the origin ;
(c) the axis is the line aj =— 4, the vertex is (— 4, 6), and the curve
passes through the point (1, 2).
13. Sketch the following parabolas and lines and find the coordinates
of their points of intersection :
(a) y = 6x%y = 'Jx-\-S. (^b) y = 2 x^ - 3x, y = x -\- 6. »
(c) y = 2-3x^,y = 2x-\-S. (d) y = S -\- x- x^, x + y - 4 = 0.
14. Sketch the following curves and find their intersections :
(a) x2 + y2 = 25, y = |x2. (&) x:^-\-y2-6y = 0,y = ^x^-2x + e.
15. The ordinate of every point of the line y :^ | a; + 4 is the sum of
the corresponding ordinates of the lines y = ^x and y = 4. Draw the last
two lines and from them construct the first line.
16. The ordinate of every point of the parabola y = lx^ + ^x— 1 is
the sum of the corresponding ordinates of the parabola y = ^x^ and the
line y = ^x — l. From this fact draw the former parabola.
17. The ordinate of every point of the parabola y = ^x^ — x + Sis the
difference of the corresponding ordinates of the parabola y = ^x^ and the
line 2/ = X — 3, In this way sketch the former parabola.
136 PLANE ANALYTIC GEOMETRY [VIII, § 134
18. Suppose the parabola y = ax^ + bx -^ c drawn, how would you
sketch the following curves ? Are these curves also parabolas ?
(a) y = a(x-\- hY + h{x + h)+c,h> 0.
(&) y = a(x- 2)2 + 5(x - 2) + c.
(c) 2/ = a(2x)2 + 6(2a;)4-c.
(d) y = a(^\xy + b(i\x)+c.
19. Find the values of x for which the following relations are true :
(a) a:2 _ a; - 12 < 0. (6) 12-a;-a:2>0.
(c) 3x2 + 6a;-2^0. {d) 5 + 13x-6x2^0.
(e) «2_5>3a; + 6. (/) x2-5<3x + 5.
20. Show that the equation of the parabola y = ax'^ -\- hx -{■ c that
passes through the points {x\ , yi), (x^ , ^2), (a^3 , yi) may be written in
the form
y x^ X \
yi
:«i2
a:i 1
2/2
3^2^
X2 1
2/3
X32
X3 1
(a) Show that if the minor of x'^ vanishes, the three given points lie on
a line.
(6) What conclusion do you draw if the minor of y vanishes ?
(c) To what does this determinant reduce if the origin is one of the
given points ?
135. Sjonmetry. Two points P^ , P^ are said to be situated
symmetrically with respect to a line Z, if I is the perpendicular
bisector of P^P^ ; this is also expressed by saying that either
point is the reflection of the other in the line I.
Any two plane figures are said to be symmetric with respect
to a line I in their plane if either figure is formed of the reflec-
tions in I of all the points of the other figure. Each figure is
then the reflection of the other in the line I. Two such figures
are evidently brought to coincidence by turning either figure
about the line I through two right angles. Thus, the lines
2/ = 2 ic -h 3 and y = — 2x — S are symmetric with respect to
the axis Ox.
VIII, §135] POLYNOMIALS— THE PARABOLA 137
A line / is called an axis of symmetry (or simply an axis) of
a figure if the portion of the figure on one side of I is the
reflection in I of the portion on the other side. Thus, any
diameter of a circle is an axis of symmetry of the circle.
What are the axes of symmetry of a square ? of a rectangle ?
of a parallelogram ?
In analytic geometry, symmetry with respect to the axes of
coordinates, and to the lines y=±x,isoi particular importance.
It is readily seen that if a figure is symmetric with respect
to both axes of coordinates, it is symmetric with respect to the
origin^ i.e. to every point Pi of the figure there exists another
point Pg of the figure such that the origin bisects PiP^. A
point of symmetry of a figure is also called center of the figure.
EXERCISES
1. Give the coordinates of the reflection of the point (a, &) in the
axis Ox ; in the axis Oy ; in the line y = X] in the line y = 2 x ; in the
line y =— X.
2. Show that when x is replaced by — x in the equation of a given
curve, we obtain the equation of the reflection of the given curve in the
y-axis.
3. Show that when x and y are replaced by y and x, respectively, in the
equation of a given curve, we obtain the equation of the reflection of the
given curve in the line y = x.
4. Sketch the lines y = — 2x + 6 and x = — 2.y -\- 5 and find their
point of intersection.
6. Sketch the parabolas y = x^ and x= y^ and find their points of in-
tersection.
6. Find the equation of the reflection of the line 2x — 3y-|-4 = 0in
the line y = x; in the axis Ox; in the axis Oy ; in the line y — —x.
7. What is the reflection of the line x = a in the line y = x? in the
axes?
8. Find and sketch the circle which is the reflection of the circle
x2 -I- y2 _ 3 ^ _ 2 = in the line y = x, and find the points in which the
two circlea intersect.
138
PLANE ANALYTIC GEOMETRY [VIII, § 135
9. Find the circle which is the reflection of the circle x^ -\-y'^ —ix +3
= in the line y = x; in the coordinate axes. Sketch all of these
circles.
10. What is the general equation of a circle which is its own reflection
in the line y = x? in the axis Ox ? in the axis Oy '? What circle is its
own reflection in all three of these lines ?
11. What is the equation of the reflection of the parabola y =—x^ + 4:
in the line y = x? in the line y = — x? Are these reflections parabolas ?
12. What is the reflection of the parabola ?/ = 3 ic'-^ — 5 x + 6 in the axis
Ox ? in the axis Oy ? Are these reflections parabolas ?
13. By drawing accurately the parabolas y -\- x^ = 1, x -{- y^ = 11, find
approximately the coordinates of their points of intersection.
14. If the Cartesian equation of a curve is not changed when x is re-
placed by — X, the curve is symmetric with respect to Oy ; if it is not
changed when y is replaced by — y, the curve is symmetric with respect
to Ox ; if it is not changed when x and y are replaced by — x and — y,
respectively, the curve is symmetric with respect to the origin ; if it is
not changed when x and y are interchanged, the curve is symmetric with
respect to y = x.
136. Slope of Secant. Let P(a;, y) be any point of the
parabola
(1) y = ax\
If Pi(xi , 2/i) be any other point of
this parabola so that
(2) 2/1 = ctx,^
the line PPi (Fig. 55) is called a
secant.
For the slope tan Oj of this secant
we have from Fig. 55 :
(3)
or, substituting for y and i/i their values :
(4) tan «i = ^W - ^') ^ a{x + x^)
Xy— X ^r — : r
SQi x^ — X Aa;
Vm, §138] POLYNOMIALS — THE PARABOLA 139
137. Slope of Tangent. Keeping the point P (Fig. 6b)
fixed, let the point Pi move along the parabola toward P; the
limiting position which the secant PP^ assumes at the instant
when Pi passes through. P is called the tangent to the parabola
at the point P.
Let us determine the slope tana of this tangent. As the
secant turns about P approaching the tangent, the point Qi ap-
proaches the point Q, and in the limit OQi = Xi becomes OQ=x.
The last formula of § 136 gives therefore tan a if we make
Xi = x:
tan a = 2
The slope of the tangent at P which indicates the '• steep-
ness " of the curve at P is also called the slojje of the parabola
at P. Thus the slope of the parabola y = ax^ at any point
whose abscissa is a; is =2 ax-, notice that it varies from point
to point, being a function of x, while the slope of a straight
line is constant all along the line.
The knowledge of the slope of a curve is of great assistance
in sketching the curve because it enables us, after locating
a number of points, to draw the tangent at each point. Thus,
for the parabola ?/ = | aj^ we find tan a = ^x ; locate the points
for which a? = 0, 1, 2, — 1, — 2, and draw the tangents at these
points ; then sketch in the curve.
138. Derivative. If we think of the ordinate of the parab-
ola y = ax^ as representing the function ax^, the slope of the
parabola represents the rate at which the function varies with
X and is called the derivative of the function ax"^. We shall
denote the derivative of y by y'. In § 137 we have proved
that the derivative of the function
y = ax^,
is y' = 2 ax.
140 PLANE ANALYTIC GEOMETRY [VIII, § 138
The process of finding the derivative of a function, which is
called differentiation, consists, according to §§ 136-137, in the
following steps : Starting with the value y= ax^ of the func-
tion for some particular value of x (say, at the point P, Fig. 55),
we give to x an increment x^—x = ^x (compare § 9) and
calculate the value of the corresponding increment y^—y^Ay
of the function. Then the derivative ?/' of the function y is the
limit that Ay / Ax approaches as Ax approaches zero. In the
case of the function y = aa^ we have
Ay=y^-y = a{x^^ - x"^) = a[(x + Axy - a;^] = a[2 xAx + (Axy^ ;
hence — = a(2 x + Ax).
Ax ^ ^ ^
The limit of the right-hand member as Ax approaches zero
gives the derivative :
y' = 2ax.-
Thus, the area y of a circle in terms of its radius x is
y = irx^.
Hence the derivative y', that is the slope of the tangent to the curve that
represents the equation y = ttx^, is
y' =2 irx.
This'represents (§ 137) the rate of increase of the area y with respect to x.
Since 2 ira; is the length of the circumference, we see that the rate of in-
crease of the area y with respect to the radius x is equal to the circumfer-
ence of the circle.
139. Derivative of General Quadratic Function. By this
process we can at once find the derivative of the general quad-
ratic function y = aa^ -{- bx -\- c (§ 131), and hence the slope of
the parabola represented by this equation. We have here
Ay = a(x + Axy -f- b(x -f- Ax) -{-c — {ax^ -\-hx-\-c)
= 2 ax Ax -\- a{Axy -\- bAx ;
hence — = 2 ax-{-b -\- aAx.
Ax
VIII, §140] POLYNOMIALS — THE PARABOLA 141
The limit, as Ax becomes zero, is 2ax-\- b; hence the deriva-
tive of the quadratic function y =.ax^ -\-hx-\- cis
y^ =2ax + h.
140. Maximum or Minimum Value. It follows both from
the definition of the derivative as the limit of Ap/Ax and from
its geometrical interpretation as the slope, tana, of the curve
that if, for any value of x, the derivative is positive, the function,
i.e. the ordinate of the curve, is {algebraically) increasing; if
the derivative is negative, the function is decreasing.
At a point where the derivative is zero the tangent to the curve
is parallel to the axis Ox. The abscissas of the points at which
the tangent is parallel to Ox can therefore be found by equat-
ing the derivative to zero. In this way we find that the
abscissa of the vertex of the parabola y = ax^ -|- 6a; -f c is
b
2a
which agrees with § 133.
We know (§ 133) that the parabola y = ax^ -\-bx-^ c opens
upward or downward according as a is > or < 0. Hence the
ordinate of the vertex is a minimum ordinate, i.e. algebraically
less than the immediately preceding and following ordinates, if
a > ; it is a maximum ordinate, i.e. algebraically greater than
the immediately preceding and following ordinates, if a < 0.
We have thus a simple method for determining the max-
imum or minimum of a quadratic function ax"^ -i-bx-^- c; the
value of X for which the function becomes greatest or least is
found by equating the derivative to zero ; the quadratic func-
tion is a maximum or a minimum for this value of x according
as a< or > 0. '
Thus, to determine the greatest rectangular area that can be inclosed
by a boundary (e.g. a fence) of given length 2 k, let one side of the
142 PLANE ANALYTIC GEOMETRY [VIII, § 140
rectangle be called x ; then the other side \^ k — x. Hence the area A of
the rectangle is
A = x{k — .r) = kx — x^.
Consequently the derivative of ^ is k — 2 x. If we set this equal to
zero, we have 2x = k, whence x — k 12. It follows that k — x — k I2\
hence the rectangle of greatest area is a square
EXERCISES
1. Locate the points of the parabola ?/ = x-^ — 4 x + | whose abscissas
are — 1, 0, 1, 2, 3, 4, draw the tangents at these points, and then sketch
in the curve.
2. Sketch the parabolas 4 y = — x'-^ + 4 x and ?/ = x^ — 3 by locating
the vertex and the intersections with Ox and drawing the tangents at
these points.
3. Is the function y = 5(x'-^ — 4 x + 3) increasing or decreasing as x
increases from x — \'> from x = | ?
4. Find the least or greatest value of the quadratic functions :
(a) 2x-2-3x + 6. (6)8-6x-x2. (c)x2-5x-5.
(d) 2-2x-x2. (e)4+x-^x'2. (/) 5 x2 - 20x + 1.
5. Find the derivative of the linear function y = mx -\- h.
6. The curve of a railroad track is represented by the equation
?/ = I x2, the axes Ox, Oy pointing east and north, respectively ; in what
direction is the train going at the points whose abscissas are 0, 1, 2, — ^ ?
7. A projectile describes the parabola y = jx—Sx^, the unit being the
mile. What is the angle of elevation of the gun ? What is the greatest .
height ? Where does the projectile strike the ground ?
8. A rectangular area is to be inclosed on three sides, the fourth side
being bounded by a straight river. If the length of the fence is a con-
stant kj what is the maximum area of the rectangle ?
9. Let e denote the error made in measuring the side of a square of
100 sq. ft. area, and E the corresponding error in the computed area.
Draw the curve representing E as ^ function of e.
10. A rectangle surmounted by a semicircle has a total perimeter of 100
inches. Draw the curve representing the total area as a function of the
radius of the semicircle. For what radius is the area greatest ?
VIII, § 143]
POLYNOMIALS
143
Fig. 56
4
18
PART II. CUBIC FUNCTION
141. The Cubic Function. A function
of the form aoX^ + a^x^ + ago; + ctg is called
a cubic function of x. The curve repre-
sented by the equation
y = aox^ -f aiO^ -I- a^x + dg
can be sketched by plotting it by points
(§ 131).
For example, to draw the curve repre-
sented by the equation
y z=: OC^ — 2 X^ — 5 X + 6,
we select a number of values of x and com-
pute the corresponding values of y :
a;=-3-2-101 2
2/=- 24 860-4
These points can then be plotted and connected by a smooth
curve which will approximately represent the curve corre-
sponding to the given equation (Fig. 56).
142. Derivative. The sketching of such a cubic curve is
again greatly facilitated by finding the derivative of the cubic
function; the determination of a few points, with their tan-
gents, will suffice to give a good general idea of the curve.
To find the derivative of the function y = aoX^ -f aiO^ + a.^
-f-ttg the process of § 138 should be followed. The student
may carry this out himself; he will find the quadratic function
y' = 3 aifii^ -f- 2 ciiX + ag-
143. Maximum or Minimum Values. The abscissas of
those points of the curve at which the tangent is parallel to
the axis Ox are again found by equating the derivative to
zero; they are therefore the roots of the quadratic equation
144 PLANE ANALYTIC GEOMETRY [VIII, § 143
3 a^ + 2 ajflj + cfca = 0.
If at such a point the derivative passes from positive to nega-
tive values, the curve is concave doiv7iivard, and the ordinate
is a maximum; if the derivative passes from negative to posi-
tive values, the curve is concave upward, and the ordinate is
a minimum.
144. Second Derivative. The derivative of a function of
X is in general again a function of x. Thus for the cubic
function y = a^T? + aiX^ + a^ + a^ the derivative is the quad-
ratic function ^f ^ 3 ^^ ^2a,x + a,.
The derivative of the first derivative is called the second deriva-
tive of the original function ; denoting it by y", we find (§ 139)
?/" = 6 (Xo^ + 2 ay.
As a positive derivative indicates an increasing function,
while a negative derivative indicates a decreasing function
(§ 140), it follows that if at any point of the curve the second
derivative is positive, the first derivative, i.e. the slope of the
curve, increases ; geometrically this evidently means that the
curve there is concave upward. Similarly, if the second de-
rivative is negative, the curve is concave downward. We have
thus a simple means of telling whether at any particular point
the curve is concave upward or downward.
It follows that at any point where the first derivative van-
ishes, the ordinate is a minimum if the second derivative is
positive ; it is a maximum if the second derivative is negative.
145. Points of Inflexion. A point at which the curve
changes from being concave downward to being concave up-
ward, or vice versa, is called a point of inflexion. At such a
point the second derivative vanishes.
Our cubic curve obviously has but one point of inflection,
viz. the point whose abscissa is ic = — ai/(3 a^).
VIII, § 145] POLYNOMIALS 145
EXERCISES
1. Find the first and second derivatives of y w^hen :
(a) ?/ = 6 x3 - 7 x2 - X + 2. (6) y = 20 + 4 x - 5 a;2 - x^.
(c) 10?/ = x3-5x2+3x + 9. {d) ?/ = (x-l)(x-2)(x-3).
(e) 2/ = x2(x + 3). (/) 7?/ = 3x-2x(x2-l).
2. Sketch the curve y = (x — 2) (x + 1) (x + 3), observing the sign of y
between the intersections with Ox, and determining the minimum, maxi-
mum, and point of inflection.
3. In the curve y = acfifi -\- aix^ + a2X + as, what is the meaning of as ?
4. Sketch the curves : »
(a) 5yz=(x-l)(x + 4)2. , (6) y=(x-3)3.
(c) 6 y = 6 + X + x2 - x8. (d) y = x^-i x.
(e) Sy = 6 x2 - x^. (/) y = x^ - 3 x2 + 4 x - 5.
5. Draw the curves y = x, y = x^, y = x^, with their tangents at the
points whose abscissas are 1 and — 1.
6. Find the equation of the tangent to the curve 14 y = 5 x^ — 2 x2
+x — 20 at the point whose abscissa is 2.
7. At what points of the curve y =x^ — ^x^ + S are the tangents
parallel to the line ?/=— 3x+5?
8. Are the following curves concave upward or downward at the
indicated points ? Sketch each of them.
(a) y = 4x3-6x, atx = 3. (b) 3y = 5x - 3 x^, at x =- 2.
(c) y = x3 - 2 x2 + 5, at X = i. (c?) 2 y = x^ - 3 x2, at x = 1.
(e) y = 1 -x-x^, atx = 0. (f) 10yz=x^+x^-l6x-{-6,a,tx=-^.
9. Show that the parabola y = ax^ +bx -^ c is concave upward or
concave downward for all values of x according as a is positive or negative.
10. The angle between two curves at a point of intersection is the
angle between their tangents. Find the angles between the curves y = x^
and y = x^ at their points of intersection.
11. Find the angle at which the parabola y = 2x2 — 3x — 5 intersects
the curve y = x^ 4- 3 x — 17 at the point (2, — 3).
12. The ordinate of every point of the curve y = x^ + 2 x2 is the sum of
the ordinates of the curves y = x^ and y = 2x^. From the latter two
curves construct the former.
L
146
PLANE ANALYTIC GEOMETRY [VIH, § 145
13. From the curve y = x^ construct the following curves :
(a) y=4:X^. {b) y = l- Y. (c) y = x^-2. (d) y = 2x^ + 4.
iij
14. Draw the curve 2y = x^ — Sx^ and its reflection in the line y = x.
What is the equation of this reflected curve ? What is the equation of
the reflection in the axis Oy ?
15. A piece of cardboard 18 inches square is used to make a box .by
cutting equal squares from the four corners and turning up the sides.
Draw the curve whose ordinates represent the volume of the box as a
function of the side of the square cut out. Find its maximum.
16. The strength of a rectangular beam cut from a log one foot in
diameter is proportional to (i.e. a constant times) the width and the
square of the depth. Find the dimensions of the strongest beam which
can be cut from the log. Draw the curve whose ordinates represent the
strength of the beam as a function of the width.
17. Show that the equation of a curve in the form y = ax^ + bx^ + ex + d
is in general determined by four points Pi (xi , yi), Po (X2 , 2/2), Ps (xs , ys),
P* (.Xi , ^4), and may be written in the form
y x^ x^ X 1
yi Xi^ xi^ xi 1
y2 X2^ X2^ X2 1
ys Xs'^ xs^ xs
= 0.
y^ Xa^ X4^ Xa
18. Find the equation of the curve in the form y = ax^ -\- bxJ^ -\- cx + d
which passes through the following points :
(a) (0,0), (2,-1), (-1,4), (3,4);
(6) (1,1), (3,-1), (0,5), (-4,1).
19. Show that every cubic curve of the form y = acfic^ + aix^ + a^x + a%
is symmetric with respect to its point of inflection.
146. Cubic Equation. The real roots of the cubic equation
a^ + a^o^ + ajOJ + ttg =
are the abscissas of the points at which the cubic curve
?y = a^"^ 4- a^Q^ + 0^2^ + ^3
intersects the axis Ox. This geometric interpretation can
VIII, § 146] POLYNOMIALS 147
be used to find the real roots of a numerical cubic equation ap-
proximately : calculating * the ordinates for a series of values
of X (as in plotting the curve by points, § 141), or at least deter-
mining the signs of these ordinates, observe where the ordinate
changes sign. At least one real root must lie between any
two values of x for which the ordinates have opposite signs.
The first approximation so obtained can then be improved by
calculating ordinates for intermediate values of x.
Thus to find the roots of the cubic
a^4-ar^-16i»-f6 =
we find that
fora; = -5 -4-3-2-10 1 2 3 4
2/ is - + + +++--- +
The roots lie therefore between — 5 and — 4, and 1, 3 and
4. To find, e.g.y the root that lies between and 1, we find that
for a; = 0.1 0.2 0.3 0.4
2/is + + + + -
The root lies therefore between 0.3 and 0.4, and as the cor-
responding values of y are 1.317 and — 0.176, the root is
somewhat less than 0.4. As
fora;= 0.40 0.39 0.38
2/ = - 0.176 - 0.029 + 0.119
a more accurate value of the root is 0.39.
This process can be carried as far as we please. But it is
very laborious. We shall see in a later section (§ 170) how
it can be systematized.
EXERCISES
1. Find to three significant figures the real roots of the equations :
(a) a;3 - 4 x2 + 6 = 0. {h) x^ + x'^ - x- \ = 0.
(c) a;3-3a:+l^=0. (d) x(x -l){x-2)=A.
* For abridged numerical multiplication and division see the note on p. 256.
148 PLANE ANALYTIC GEOMETRY [VIII, § 147
PART III. THE GENERAL POLYNOMIAL
147. Polynomials. The methods used in studying the
quadratic and cubic functions and the curves represented by
them can readily be extended to the general case of the poly-
nomial, or rational integral function, of the nth degree,
y = a^x" + a^sf"-^ + a^vf"-^ H \- a^_^x + a„ ,
where the coefficients «„, a^, ••• a„ may be any real numbers,
while the exponent n, which is called the degree of the poly-
nomial, is a positive integer.
We shall often denote such a polynomial by the letter y or
by the symbol f{x) (read : function of x, or / of a;) ; its value
for any particular value of x, say x = Xy or x — h, is then de-
noted by /(iCi) or fQi), respectively. Thus, for x = we have
/(0) = a,.
148. Calculation of Values of a Poljmomial. In plotting
the curve y=f(x) by points (§§ 131, 141) we have to calculate
a number of ordinates. Unless f{x) is a very simple poly-
nomial this is a rather laborious process. To shorten it ob-
serve that the value /(x^) of the polynomial
f(x) = aox"" -}- ai«"-i -f • . • -\-a^
for x = Xi can be written in the form
f{xi) =( ... {((aoXi + ai)xi-\-a2)Xi-i-as)Xi-\- \- a^_{)x-\- a,.
To calculate this expression begin by finding aQX^ -f a^ ; mul-
tiply by Xy and add a^ ', multiply the result by x^ and add a^ ;
etc. This is best carried out in the following form :
Oo % ttg • • • ct,t
Opa^ (a^Xy -f g^) x^
a^flOi -f ai (a^Xi -f- ai)Xi + ^g • • •
Eor instance, if
f(x) = 2 a^ - 3 «2 _ 12 a; + 5
= ((2a:-3)a;-12)a;-f-5,
Vm, § 149] POLYNOMIALS 149
to find /(3) write the coefficients in a row and place 2x3 = 6
below the second coefficient ; the sum is 3. Place 3 x 3 = 9 be-
low the third coefficient ; the sum is — 3. Place 3x(— 3)= — 9
below the last coefficient; the sum, —4, is =/(3).
2-3-12 5
6 9 -9
2 3 _ 3 _4
This process is useful in calculating the values of y that cor-
respond to various values of x, as we have to do in plotting a
curve by points. It is also very convenient in solving an equa-
tion by the method of § 146.
EXERCISES
1. If /(«)= 5x3 _ ISx + 2, what is meant by /(a)? by f{x + A) ?
What is the value of /(O)? of /(2) ? of /(- 3 5)? of/(-l)?
2. Find the ordinates of the curve y = x* — x^ + 3 x^ — 12 x + 3 for
X = 3, - 9, - i.
3. Find the ordinates of 2 y = x* + 3 x^ - 20 x - 25 for x = 1, 2, 3, - 1,
-2. ■
4. Suppose the curve y =/(x) drawn ; how would you sketch :
(a) 2/=/(x-2)? {b) y = f(x+S)? (c) y = f(2x)? (d) y=f(-x)?
(e) y=f(^^y if)y=f(x)+5? (g) y =f(x-)-2x?
6. Calculate to three places of decimals the real roots of -the equations :
(a) x3 -f x2 = 100 ; (&) x^ - 4 = ; (c) x^ - 7x + 7 = 0.
149. Derivative of the Polynomial. We have seen in the
preceding sections how greatly the sketching of a curve and
the investigation of a function is facilitated by the use of the
derivatives of the function. Thus, in particular, the first
derivative y' is the rate of change of the function y with x,
and hence determines the slope, or steepness, of the curve
y =f{x). We begin therefore the study of the polynomial by
determining its derivative. The method is essentially the
150
PLANE ANALYTIC GEOMETRY [VIII, § 149
same as that used in §§ 138, 139 for finding the derivative of
a quadratic function.
The first derivative y' of any function 2/ of a; is defined, as
in § 138, to be the limit of the quotient Ay/ Ax as Aa; approaches
zero, Ay being the increment of
the function y corresponding to
the increment Ax of a? ; in symbols :
y'
lim^,
Ax=o Aa;
y
y^ A 1 1
^
// 1 I
M 1 1
X
^
/ N .Q,
A
Fig. 57
Geometrically this means that y'
is the slope of the tangent of the
curve whose ordinate is y. For, Ay/ Ax is the slope of the secant
PP, (Fig. 57) :
— ^ = tan «! ;
Ax '
and the limit of this quotient as Aa; approaches zero, i.e. as P^
moves along the curve to P, is the slope of the tangent at P:
y' = tan a = lim ~
Aw
im -^
Ax=o Aa;
If the function y be denoted by /(«), then
Ay=f(x-\-Ax)-f(x)',
hence
y
^.^./XaH^A^WM^
Aa^ Aa;
150. Calculation of the Derivative. To find, by means of
the last formula, the derivative of the polynomial
y =f(^) = cto*" + «!«?"-' + • • • + a„,
we should have to form first /(a; 4- Aa;), i.e.
(x + Axy + a,{x-hAxy-^-\- ... -fa„,
subtract from this the original polynomial, then divide by Aa;,
and finally put Aa; = 0.
VIII, § 151] POLYNOMIALS • 151
This rather cumbersome process can be avoided if we
observe that a polynomial is a sum of terms of the form ax^
and apply the following fundamental propositions about
derivatives :
(1) the derivative of a sum of terms is the sum of the deriva-
tives of the terms ;
(2) the derivative of ax"" is a times the derivative of x"" ;
(3) the derivative of a constant is zero;
(4) the derivative of x"" is nx'^'K
The first three of these propositions can be regarded as
obvious ; a fuller discussion of them, based on an exact defi-
nition of the limit of a function, is given in the differential
calculus. A proof of the fourth proposition is given in the
next article.
On the basis of these propositions we find at once that the
derivative of the polynomial
y = ao.T" -f- aiO?""^ -f OoX'^'^ 4- . . . -f- a,,^iX + a„
is
y' = ao«a?"-^ + «! {n — 1) a;"-^ + ag {n — 2)a7"-^ 4- • • • + a„_i •
151. Derivative of ic»*. By the definition of the derivative
(§ 149) we have for the derivative of y = x'':
Aa^=0 Ax
Now by the binomial theorem (see below, § 152) we have
(x -f Axy = .T'» -h nx^'-'^Ax + ^^i'^ — ^) a;»-2(Ax)- + ... -f (Aa;)%
and hence
{x -f Axy - X" = nx''-^Ax-\- ^(^' ~ -*^) x"-2(Aic)2 -|- ... -f (Ax)\
Dividing by Ax and then letting Ax become zero, we find
y' =z 7ia;"~\
152 PLANE ANALYTIC GEOMETRY [VIII, § 151
EXERCISES
1. Find the derivatives of the following functions of x by means of
the fundamental definition (§ 149) and check by § 150 :
(a) x\ (b) x^ + x. (c) x^+6x-^.
(d) -6x3. (e) a;* -3x3. (y) rnx -\- b.
2. Find the derivatives of the following functions :
(a) 5x4-3x2 + 6x. (6) l-x + ^x^-^x^ (c) (x-2)3.
(d) (2x + 3)5. (e) 3(4^ X- 1)3. (/) x'»+ax"-i + 6x"-2.
3. For the following functions write the derivative indicated :
(a) 5 x3 - 3 X, find y"'. (&) ax^ + 6x + c, find y'".
(c) x6, find y\ (^) «^^ + ^^^ + ex + (Z, find y''\
(e) ix6, find y". (/) ^a;6, find ?/-».
(g) xi2 _ g^xs, find y'". (h) (2 x - 3)^, find y'".
152. Binomial Theorem, in § 151 we have used the binomial
theorem for a positive integral exponent w, i.e. the proposition that
(1) (x + A)" = x« + wa;«-iA + Vlllsi^x»-%^ + ^(»-l)(y^-2) ^.n-g^^a
+ ... + ^iiLrA)-Ah-,
n !
The formula (1) can be proved by mathematical induction (§ 62). It
certainly holds for n = 2, since by direct multiplication we have
(X + /i)2 = x2 + 2x^1 + A2 = x2 + 2x^ + ^h^,
21
which agrees with (1) for n = 2.
Moreover, if the formula (1) holds for any exponent w, it holds for
n + 1. For, multiplying (1) by x -\- h in both members, we find
(X + h)^+^ =
xn+i + (n + l)xnh + (^ + ^^^ X-1A2 + ... + (n + l)n{n -1) .•• 1 ^^^^^
which is the form that (1) assumes when n is replaced by n + 1.
153. Binomial Coefficients. The coefficients
2 ! ' (n-l)l ' w !
in the binomial formula (1) are called the binomial coefficients.
VIII, § 153] POLYNOMIALS ' 153
The meaning of these coefficients will appear from another proof of the
formula, which is as follows : If n is a positive integer, we can write
(x + y^)" in the form
(x + hi){x + h2){x + hz){x + hi) ... {X + /i„),
where the subscripts are used simply for convenience to distinguish the
binomial factors ; i.e. it is understood that hi = h2 = hz= ••• = hn= h.
Each term in the expanded product is the product of n letters of which
one and only one is taken from each binomial factor. To form all these
terms we may proceed as follows :
(a) If we choose x from each of the n factors, we obtain as first term
of the expansion x^.
(b) If we choose x from n — 1 factors, the letter h can be chosen from
any one of the n factors, i.e. it can be chosen in „(7i ways (§ 64) ; this
gives
x'*-'^(hi 4- ^2 + ••• + ^n)i the number of terms being „Ci.
(c) If we choose x from n — 2 factors, the other two letters can be
chosen from any two of the n factors, i.e. in ^(h ways ; this gives
x^~^(hih2 + hihs 4- ••• + ^2^3 + •••)» ^^^ number of terms being „CV
(d) If we choose x from n — 3 factors, the other three letters can be
chosen from any three of the n factors, i.e. in nCs ways ; this gives
x>'-^(hihihz-\-hxh2h^ + ••• +^2^3^4 + •-•)» ^^^ number of terms being ^Cg.
Finally we have to choose no x and consequently an h from every factor,
which can be done in „C„=1 way ; this gives the last term
hihi -' K.
Now as ^1 = A2 = ••• =h„=h, we find the binomial expansion :
(X + h)^ = a:« + nCiX'*-!^ + rtCiX^-^h^ + ••• + nCn-lXh^'^ + nCrM.
Since, by § 64,
1 • Ji
this form agrees with that of § 152. It will now be clear why the
binomial coefficients are the numbers of combinations of n elements,
1, 2, 3, ••• at a time.
154 PLANE ANALYTIC GEOMETRY [VIII, § 153
The proof also shows that the binomial coefficients are equal in pairs,
the first being equal to the last, the second to the last but one, etc.
Finally it may be noted that, with cc = 1, ^ = 1 we obtain the following
remarkable expression for the sum of the binomial coefficients :
EXERCISES
1. Show that in the expansion of (x— ^)'* by the binomial theorem the
signs of the terms are alternately + and — .
2. If the binomial coefficients of the first, sec- 1
ond, third, fourth, etc., power of a binomial are 1 1
12 1
written down as in the horizontal lines of the
adjoining diagram, it will be observed that (ex- 14 6 4 1
cepting the ones) every figure is the sum of the 1 5 10 10 5 1
two just above it. Extend the triangle by this rule
to the 10th power, and prove the rule (see § 152). Pascal's Triangle
3. Expand by tne binomial theorem :
(a) {x + 2y)^ ' (6) (^'+1)'- (c) (2a-c)3.
00 (--^X- («) (a + b + cy. (/) {4:x-lyy.
\y x^J
(g) (H-2x)3-(l-2x)3. (^h) (l^xy^ (0 (^-^)*'-
(.0 (f-^-l)'- W (|--a^2)*. (0 (a + b-c-dy.
4. Write the term indicated :
(a) Fourth term in (a 4- by^. (d) Middle term in (x^ — y^y^.
(6) Fourth term in (a - by^. (e) A;th term in (x + hy.
(c) Tenth term in (x^ -f 4 y^y^. (/) kth term in (x - hy.
(g) Two middle terms in (a^ _ 2 b'^y.
(1 \2'^
a — ] .
5. Show that the sum of the coefficients in the expansion of {x—hy is
zero when n is an odd integer.
6. Use the binomial formula to find {a) (1.02)6; (5) (3.97)«.
VIII, § 155] POLYNOMIALS 155
154. Properties of the General Polynomial Curve. In plot-
ting the curve
y = OoX" 4- ajic""^ -h a2^"~^ H- •• • + «„
observe that (Fig. 58) :
(a) the intercept OB on the axis Oy
is equal to the constant term a„ ;
(h) the intercepts OA^, OA^, •••on
the axis Ox are roots of the equation
y = 0, i.e. ' Fig. 58
(c) the abscissas of the least and greatest ordinates are
found by solving the equation y' = 0, i.e. (§ 150)
every real root giving a minimum ordinate if for this root y"
is positive and a maximum ordinate if y" is negative ;
(d) the abscissas of the points of inflection are found by
solving the equation y" =^0, i.e.
n(w-l)ao^"-2+ ... +2a„_2 = 0,
every real root of this equation being the abscissa of a point
of inflection provided that y"'=^0. (If y'" were zero, y' might
not be a maximum or minimum, and further investigation
would be necessary.)
155. Continuity of Polynomials. It should also be ob-
served that the function y = a^pf + Oja;""^ + — + a„ is one-
valued, real, and finite for every x ; i.e. to every real and finite
abscissa x belongs one and only one ordinate, and this ordinate is
real and finite. Moreover, as the first derivative y' = noojc""^
+ ••• +«„_! is again a polynomial, the slope of the curve is
everywhere one-valued and finite.
156
PLANE ANALYTIC GEOMETRY [VIII, § 155
Thus, so-called discontinuities of the ordinate (Fig. 59) or of
the slope (Fig. 60) cannot occur : the curve y = a^'' 4- — -\- a„
is continuous.
Strictly defined, the continuity of the function y = a^"" + —
4- a„ means that, for every value of x, the limit of the function
is equal to the value of the function. The function y = a^fid^ + •••
4-a„ has one and only one value for any value x = x^j viz.
(^1 -\- •'• +^n- The value of the function for any other
value of X, say for oci + Ace, is a^i(Xl -f Aa?)'* + — + a„ which can
be written 'in the form aQXi" -\- — +a„-l- terms containing Aa;
as factor. Therefore as Aa; approaches zero, the function
approaches a limit, viz. its vahie for x=:Xi.
156. Intermediate Values. A continuous function, in
varying from any value to any other value, must necessarily
pass through all intermediate values. Thus, our polynomial
y = a(fic'' -f ..• -f a„, if it passes from a negative to a positive
value (or vice versa), must pass through zero. It follows
from this that heticeen any two ordinates of opposite sign the
curve y = aoX'* + ••• + ^„ must cross the axis Ox at least once.
It also follows from the continuity of the polynomial and
its derivatives that between any two intersections ivith the axis
Ox there must lie at least one maximum or minimum^ and be-
tween a maximum and a minimum there must lie a point of
inflection.
Ordinates at particular points can be calculated by the pro-
cess of § 148.
VIII, § 156] POLYNOMIALS 157
EXERCISES
1. Sketch the following curves :
(a) y=(x-l)ix-2){x-S). (&) 4i/ = x4-l. (c) 10y = x^.
(d) 10y = x^-\-5. (e) iy={x-h2)%x^S). (/) y={x-lY.
2. When is the curve y = aox** + aix»-i + •••+«« symmetric with
respect to Oy ?
3. Determine the coefficients so that the curve y = aox'^ + aix^ + azx^
+ a^x + a4 shall touch Ox at (1, 0) and at (— 1, 0) and pass through
(0, 1), and sketch the curve.
4. Find the coordinates of the maxima, minima, and points of inflec-
tion and then sketch the curve 4 ^ = x"* — 2 x^.
5. Are the following curves concave upward or downward at the indi-
cated points ?
(a) 16^ = 16x4-8x2 + 1, atx=-l, - |, 0, |, 3.
■ (6) y=4:X-xS at X = - 2, 0, 1, 3.
(c) y = X", at any point ; distinguish the cases when n is a positive
even or odd integer.
6. What happens to the curves y = ax^ and y = ax^ as a changes ?
For example, take a = 2, 1, ^, 0, — |, — 1, — 2.
7. Find the values of x for which the following relations are true :
(a) x* - 6 x2 + 9 ^ 0. (6) (x - l)2(x2 - 4) ^ 0.
8. Show that the following curves do not cross the axis Ox outside of
the intervals indicated :
(a) ?/ = x^ — 2 x2 4- 4x + 5, between — 2 and 2.
(&) i/ = x4-5x2 + 6x-3, -3 and 3.
(c) y = x3-x2 + 3x-3, Oand 1.
(d) y = x4 + x2 - 3 X + 2, and 1.
9. Those curves whose ordinates represent the values of the first,
second, etc., derivatives of a given polynomial are called the first, second,
etc., derived curves. Sketch on the same coordinate axes the following
curves and their derived curves :
(a) 6^=2x3-3x2- 12x. (b) y = (x- 2y(x + 1).
(c) y = (x+ 1)3. (d) 2 y = X* + x2 + 1.
10. At what point on Ox must the origin be taken in order that the
equation of the curve y = 2x^ — Sx^ — 12x — 5 shall have no term in x2 ?
no term in x ?
158 PLANE ANALYTIC GEOMETRY [VHI, § 157
PART IV. NUMERICAL EQUATIONS
157. Equations. Roots. In plotting the curves y — afpif +
••• + ^n (§ ^^^) i* is often desirable to solve equations of the form
(1) ao^;" + - + «„ = 0,
the coefficients «o> %> ••* «« being given real numbers and n any
positive integer. The solution of such numerical equations^
at least approximately, presents itself in many other prob-
lems. The roots of the equation (1) are also called the roots,
OY zeros, of the function a^fc'' -\- ••• +ot„-
It is understood that a^ 4^ since otherwise the equation
would not be of degree n. We can therefore divide (1) by a^
and write the equation in the form
(2) x^+x>^:ff-^-^ ... H-i),=0,
where p^=ia^/aQ, i>2=«2/«o? — i>,» = «„/^o are given real numbers.
158. Relation of Coefficients to Roots. A\^e here assume
the fundamental theorem of algebra that every equation of the
form (2) has at least one root, say x = x^, which may be real or
imaginary. If we then divide the polynomial x^-\-p^x''~^-\ \-p^
by a; — Xy, we bbtain a polynomial of degree ii — 1 ; the equation
of the (n — l)th degree obtained by equating this polynomial to
zero must again have at least one root. Proceeding in this
way, we find that every equation of the form (2) has n roots,
which of course may be real or complex, and some of which
may be equal. It also appears that the equation (2) may be
written in the form
(3) {x - x^){x -x^"'{x- a;„) = 0,
where x^, x^, •.« x^ are the n roots, or performing the multiplica-
tion (§ 153) :
(4) a;" - (a?! + . . . -f x„)a;"-i -f- {x^x. + • • • + x^^^x^x"""^ -f • • .
-|-(— 1)% ... 05^ = 0.
VIII, § 159] NUMERICAL EQUATIONS 159
Comparing the coefficients in (4) with those in (2), we find :
Xi + -"+Xn = -pi,
i.e. if the coefficient of the highest power of a polynomial is
one, then the coefficient of a;"~^, with sign reversed, is equal to
the sum of the roots; the coefficient of x""'^ is equal to the sum
of the products of the roots two at a time ; minus the coefficient
of «'*"' is equal to the sum of the products of the roots three at
a time, etc. ; plus or minus the constant term (according as n is
even or odd) is equal to the product of all the roots.
159. Equations with Integral Coefficients. The results of
the last article can often be used to advantage to find the roots
of a numerical equation (2) in which all the coefficients pi, "-p^
are integers. We then try to resolve the left-hand member
into linear factors of the form x — x,^; if this can be done, the
roots are the numbers x^.
The fact that the constant term j)^ in (2) is plus or minus
the product of the roots can be used in the same case by trying
to see whether any one of the integral factors of ± p^ satisfies
the equation.
EXERCISES
1. Findtherootsof : (a) x^ - 7 a;-}-6=0 ; (&) a;3-2 x2-13a;-10=0 ;
(c) x* - 1 = ; (d) x* - 7 x2 - 18 = ; (e) a;^ - 5 a;2 - 2 x + 24 = 0.
2. Form the equation whose roots are : (a) 2, — 2, 3 ; (&) — 1, — 1, 1 ;
(c) 0, V2, -V2; (d) -1, 1, i, -f.
3. For the equation x^ + pix^ + pox -\-p3 = determine the relation
between the coefficients when : (a) two roots are equal but opposite in
sign ; (6) the product of two roots is equal to the square of the third ;
(c) the three roots are equal.
4. Show that the sum of the n nth roots of any number is zero. What
about the sum of the products of the roots two at a time '? three at a time ?
160 PLANE ANALYTIC GEOMETRY [VIII, § 160
160. Imaginary Roots. In general, the real roots of a
numerical equation are of course not integers, nor even rational
fractions, but irrational numbers. In solving such an equation
the object is to find a number of decimal places of each root
sufficient for the problem in hand. Methods of approximation
appropriate for this purpose are given in the following articles.
The imaginary roots of the equation can be determined by
somewhat similar, though more laborious, processes. It will
here suffice to show that imaginary roots always occur in pairs
of conjugates ; that is, if an imaginary number a -\- pi is a root
of the equation (1) (with real coefficients), theii the conjugate
imaginary number a — /Si is a root of the same equation.
For, substituting a -j- ^i for x in (1) and collecting the real
and pure imaginary terms separately, we obtain an equation of
the form A-{-Bi = 0,
where A and B are real ; hence, by § 116, ^ = and B = 0.
If, on the other hand, we substitute in (1) a — /3^ for x, the
result must be the same except that i is replaced by — i; we
find therefore A — Bi = 0, and this is satisfied if A = and
^ = 0, i.e. if a -h (ii is a root.
It follows in particular that a cubic equation always has at
least one real root. Indeed, in the case of the cubic equation,
only two cases are possible : (a) the equation has three real
roots, which may of course be all different, or two equal but
different from the third, or all three equal ; (b) the equation
has one real and two conjugate imaginary roots.
161. Methods of Approximation for Real Roots. If a good
sketch of the curve y = aQX''-\-, ••• + a„ were given, we could
obtain approximate values of the real roots of the equation
ao'K" 4- ••• +a^ =
by measuring the intercepts OA^j OA^, etc., made by the curve
VIII, § 162]
NUMERICAL EQUATIONS
161
on the axis Ox (§ 154). If the curve is not given, we calculate
a number of ordinates for various values of x until we find
two ordinates of opposite sign ; we know (§ 156) that the curve
must cross the axis Ox between these ordinates, and therefore
at least one real root of the equation must lie between the
abscissas, say x^ and x^, whose ordinates are of opposite sign.
We can next contract the interval between which the root
lies by calculating intermediate ordinates. By this process a
root can be calculated to any desired degree of accuracy. But
the process is rather long and laborious. The calculation of
the ordinates is best performed by the process of § 148.
162. Interpolation. If the interval within which the root
has been confined is small, we can obtain, without calculating
further ordinates, a further approximation to the root by
replacing the curve in the interval by its secant, and finding
its intersection with the axis Ox.
Suppose (Fig. 61) that we have
found that a root lies between
OQy = Xy and 0Q2 = X2, the ordi-
nates QiPi = 2/1 and Q2P2 = 2/2 being
of opposite sign. Then Xi is ^. first
approximation to the root a;; and
if Qi and Q2 lie close together, the
intercept OQ made by the secant
PjPo on the axis Ox is a second approximation. Let us
calculate the correction Q^Q = h which must be added to
the first approximation x^ to obtain the second approximation
x^ + h.
The figure shows that QiQ/BP2 = PiQi/PiR, i-e.
Fig. 61
2/2-2/1
162 PLANE ANALYTIC GEOMETRY [VIII, § 162
hence the correction li is
7i =
Out) — ~ tJC-i LAtJu
^ 2/1 = - —2/1-
Vi - Vi ^y
This process, which is the same as that used in interpolating
in a table of logarithms, is known as the regula falsi, or rule
of false position.
163. Tangent Method. Another method for finding a correction
consists in using the intercept made on the axis Ox not by the secant but
by the tangent to the curve at Pi.
The correction Qi Q' = k is found
(Fig. 61) from the triangle PiQiQ', in
which the tangent of the angle at Q' is
equal to the value of the derivative yi'
at Pi. This triangle gives
k
hence k=— ^- .
y\'
yi .
k'
Fig. 61
Find by this method the roots of 0^ — 305+1=0.
164. Newton's Method of Approximation. After finding,
by § 161, a first approximation x^ to a root of the equation
(1) ao^" + a,x--' + . . . + a„ = 0,
transfer the origin to the point (xi, 0). Thus (Fig. 62), if a
root lies between 3 and 4, transform the
equation to (3, 0) as origin, by replacing
a; by 3 -h h. An expeditious process for
finding the new equation in h, say
(2) 6o/i«4-M"-' + ••• + &„= 0,
will be given in §§165-167.
Fig. 62
VIII, § 166] NUMERICAL EQUATIONS 163
As li is a proper fraction, its higher powers will be small,
so that an approximate value of h can be obtained from the
linear terms, i.e. by solving h,,_^h-\-h,^ =0, which gives h ap-
proximately = — 6„ / 6„_i. Hence we put
(3) h = -^-^k,
where A: is a still smaller proper fraction. If the approxima-
tion obtained from the linear terms should be too rough, we
may find a better approximation of h by solving the quadratic
K-2h'+K_,h + K = 0.
We next substitute the value (3) of h in (2) and proceed in
the same way with the equation in k. The process can be
repeated as often as desired ; the last division can be carried
to about as many more significant figures as have been obtained
before. The example in § 168 will best explain the work.
165. Remainder Theorem, if a polynomial f(x) = aox» +
aix^-^ 4- • . • + a„ of degree n be divided by x — h, there is obtained in
general a quotient Q, which is a polynomial of degree n — I, and a re-
mainder B :
^= Q + ^, i.e. fix) = Q(x-h) + B.
X — h X — h
For X = h the last equation gives /(^) = B ; i.e. the value of the poly-
nomial for any particular value h of x is equal to the remainder B ob-
tained upon dividing the polynomial by x — h :
fih) = aoh^ + -'--\-an = B.
This proposition is known as the remainder theorem.
166. Synthetic Division. As an example let us divide
f(x) = 2 x3 - 3 ic2 - 12 a; -I- 5
"by a; — 3. By any method we obtain the following result :
X — 3 x— S
164 PLANE ANALYTIC GEOMETRY [VIII, § 166
The elementary method is as follows :
2 a;8 - 6 x^
3 x2 - 12 X
3a;2_ 9a;
-3x + 5
- 3 X + 9
-4
This process can be notably shortened :
(a) As the dividend is a polynomial, it can be indicated sufficiently by
writing down its coefficients only, any missing term being supplied by a
zero ; 2 - 3 - 12 5
(5) As X in the divisor has the coefficient 1, the first terms of the
partial products need not be written ; the second terms it is more con-
venient to change in sign ; in other words, instead of multiplying by — 3
and subtracting, multiply by + 3 and add.
The whole calculation then reduces to the following scheme :
2-3-12 5[3
6 9-9
2 3-3-4
This is the same scheme as that in § 148. But it should be observed
that this method, known as synthetic division, gives not only the remain-
der — 4, i.e. /(3), but also the coefficients 2, 3, — 3 of the quotient.
167. Calculation of /(aj^ + h). if in /(x) = aox« + • • • + «„ we sub-
stitute x = Xi-{-h, we find :
/(x) =/(xi + h) = ao(xi + h)^ + ai(xi + A)«-i + ••. + a«-i (xi + h) + a„.
Expanding the powers of Xi 4- A by the binomial theorem and arrang-
ing in descending powers of h we obtain a result of the form
/(x) =/(xi + h)=b^^ + hih^-^ + ••• 4- bn.ih + 6„.
To find the coefficients 6o, 6i,--- &„ of this expansion of /(xi + h) in
powers of h observe that as A = x — Xi we have
/(x) =/(xi + h) = bo{x - xi)« 4- l>i(x -^ xi)"-! + ••. + 6„_i(x - xi) + &«.
The last term, 6„, is therefore the remainder obtained upon dividing
f(x) by X— xi ; it is best found by synthetic division (§ 166). The quo-
tient obtained upon dividing /(x) by x — Xi is evidently 6o(x — xi)""^
+ &i(x — xi)''-^ 4- ••• 4 6„_i ; the last term, 6„_i, can again be obtained
VIII, § 168] NUMERICAL EQUATIONS 165
as the remainder upon division by x — xi. Proceeding in this way all
the coefiBcients 6„, b^-u ••• &i, &o can be found.
For the example of § 166 we have
2-3-12 5|3
6
9
-9
2
3
-3
-4
6
27
2
9
6
24
2 15
The result is : f{S-{-h)=2h^ + 15h'^ + 2^h- 4.
168. Example. The roots of the equation
2 a;3 _ 3 a;2 _ 12 X + 5 =
are readily found to lie between — 3 and — 2, and 1, 3 and 4. To
calculate the last of these we find by transferring the origin to the point
(3, 0) the following equation for the correction h to the first approxima-
tion, which is 3 (§ 167) :
The linear terms give h = 1/6 = 0.17; as the quadratic term, 15 h^, is
about 0.42 and 1/24 of this is 0.02, a somewhat better approximation is
h = 0.15. Substituting
Ji = 0.16 + hi,
we find: 2 15 24 -4
0.30 2.295 3.94425
2 15.30 26.295 -0.05575
.30 2.340
2 15.60 28.635
^
2 15.90
Hence the equation for hi is
2 hi^ + 15.90 hi^ + 28.635 hi - 0.06575 = 0.
The linear terms give ^ = 001947
As the quadratic term can influence only the 6th decimal place, we can
certainly take ^1 = 0.00195 and thus find the root 3.15195.
166 PLANE ANALYTIC GEOMETRY [VIII, § 169
169. Negative Roots. To find a negative root replace a; by — x
in the given equation, i.e. reflect the curve in the axis Oy.
To find a root greater than 10 replace x by 10 «, or 100 2;, etc., in the
given equation, and calculate z.
170. Horner's Process. W. G. Homer's method is essentially the
same as Newton's, inasmuch as it consists in moving the origin closer
and closer up to the root. But it calculates each significant figure
separately. Thus, for the example of § 168 we should proceed as follows:
As in §§ 167, 168, we diminish the roots of the equation
2x3-3x2_i2x + 5 =
by 3 so that the equation (as there shown) takes the form
2 a;3 + 15 a:2 + 24 X - 4 = 0.
The left-hand member changes sign between 0.1 and 0.2. We move there-
fore the origin through 0.1 to the right :
2 15 24 -4
.2 1.52 2.552
2 15.2 25.52 -1.448
.2 1.54
2 16.4 27.06
.2
2 15.6
The new equation is 2 x^ + 15.6 x^ + 27.06 x - 1.448 = 0.
The left-hand member changes sign between 0.05 and 0.06 ; hence we
move the origin through 0.05 :
2 15.6 27.06 -1.448
.10 .785 1.39225
2 15.70 27.845 -0.05576
.10 .790
16.80 28.636
.10
2 15.90
The new equation is 2 x^ + 15.90 x^ + 28.636 x - 0.05575 = 0.
We can evidently go on in the same way finding more decimal places.
It should not be forgotten (§ 164) that after finding a number of significant
VIII, § 170] NUMERICAL EQUATIONS 167
figures in this way, about as many more can be found by simple division.
Thus, we have found x = 3.15 ••• ; the linear terms of the last equation
give the correction 0.00195, so that x = 3.15195.
EXERCISES
1. Find: (a) the cube root of 67 ; (&) the fourth root of 19 ; (c) the
fifth root of 7, to seven significant figures, and check by logarithms,
2. Newton used his method to approximate the positive root of
x^ — 2ic — 5=0; find this root to eight significant figures.
3. Find, to five significant figures, the root of the equation
x^ + 2.73 x^ = 0.375.
4. Find the coordinates of the intersections of the curve y
= {x- l)2(x+2) with the lines : (a) y = S; (6) y=l x+1; (c) y=}x-l.
5. After cutting off slices of thickness 1 in., 1 in., 2 in., parallel to
three perpendicular faces of a cube, the volume is 8 cu. in. What was
the length of an edge of the cube ?
6. Find the radius of that sphere whose volume is decreased 50%
when the radius is decreased 2 ft.
7. For what values of k will the lines kx + y + 2 = 0, x -\- ky — 1 = 0,
2x— y -{- k = pass through a common point ?
8. For what values of k are the following equations satisfied by other
values of x, y, z, to than 0, 0, 0, 0? kx + 2y-{-z — Sw = 0, 2x +ky + z
— w = 0, X— 2y-^kz-{-w = 0, x+7y — z + kw=0-
9. A buoy composed of a cone of altitude 6 ft. surmounted by a
hemisphere with the same base when submerged displaces a volume of
water equal to a sphere of radius 6 ft. Find the radius of the buoy.
10. Find, to four significant figures, the coordinates of the intersections
of the parabolas y -{- x^ = 7, x + y^ = U, Ex. 13, p. 138.
11. By applying Newton's method (§ 164) to both coordinate axes
simultaneously, find that intersection of the parabolas x^ — y = 4 and
X + y2 — 3 which lies in the first quadrant.
12. The segment cut out of a sphere of radius a by a plane through
its center and a parallel plane at the distance x from it has a volume
= irx(a:^ — iaj2); at what distance from its base must a hemisphere be
cut by a plane parallel to the base to bisect the volume of the hemisphere ?
168 PLANE ANALYTIC GEOMETRY [VIH, § 171
171. Expansion of f{x + h). The solution of numerical equa-
tions is based on the fundamental fact (§ 167) that if f(x) is a poly-
nomial, then f(x\ f K) can be expressed as a polynomial of the same
degree in A, and the coefficients Aq, J.i , ••• J.„ of this polynomial can be
calculated. Thus, for
f(x) —a^p^ + aix^ + a-ix^ + a^x + a^
we have :
/(i»i+ h) = «o (xi + hy + ai (xi + hy + aa (xi + hy+ as (xi + h)+a4
= «o^i* + <^i^i^ + «2a;i^ + asXi + a4
+ (4 aoXi^ + 3 aiXi^ + 2 a^xi + as)^
+ (6 aoXi2 + 3 aiXi + aa)^"^
+ (4aoXi + ai)/i3
+ ao^*.
Now this process is closely connected with that of finding the successive
derivatives of the polynomial. Thus we have for
f{x) = uqX'^ + aix^ + a20c^ + asx + at
the derivatives :
f'{x) = 4 aoSc8 + 3 aix^ + 2a2X + as,
f"lx) = 12 aoic2 + 6 aix + 2 a2,
/'"(x) = 24aoa; + 6ai,
/-(x)=24ao,
all higher derivatives being zero. If in these expressions we put x = xi
and then multiply them respectively by 1 , /i, h^/2 ! , h^/S ! , h*/4: ! , and
add, we find precisely the above expression for /(xi + h); hence we have:
whenever /(x) is a polynomial of degree 4.
It can be proved in the same way that for a polynomial of degree n
we have
f(xi + h)=f(xO + f'ix^)h+^-^^h'^+--'+-^^^^^
This formula is a particular case of a general proposition of the differ-
ential calculus, known as Taylofs theorem. It shows that the value of a
polynomial for any value x = Xi -\- h can he found if we know the value of
the polynomial itself and of all its n derivatives for some particular
value xi of x. This property is characteristic for polynomials.
CHAPTER IX
THE PARABOLA
172. The Parabola. The parabola can be defined as the
locus of a point whose distance from a fixed point is equal to its
distance from a fixed line. The fixed point is called the focus,
the fixed line the directrix, of the parabola.
Let F (Fig. 63) be the fixed point, d the fixed line ; then
every point P of the parabola must satisfy
the condition
FP=:PQ,
Q being the foot of the perpendicular from
P to d. Let us take F as origin, or pole, and
the perpendicular FD from jP to the directrix
as polar axis, and let the given distance FD
= 2 a. Then FP =r and PQ = 2 a —r cos cf>.
The condition FP = PQ becomes therefore
r = 2 a — r cos cf>,
2a
I.e.
(1)
1 -f cos <|>
This equation, which expresses the radius vector of P as a
function of the vectorial angle </>, is the polar equation of the
parabola, when the focus is taken as pole and the perpendicular
from the focus to the directrix as polar axis.
173. Polar Construction of Parabolas. By means of the
equation (1) the parabola can be plotted by points. Thus, for
<^ = we find r = a as intercept on the polar axis. As <^
increases from the value 0, r continually increases, reaching
169
170
PLANE ANALYTIC GEOMETRY [IX, § 173
the value 2 a for <^ = i tt, and becoming infinite as <f> ap-
proaches the vahie tt.
For any negative value of <^ (between and — tt) the radius
vector has the same length as for the corresponding positive
value of <^ ; this means that the parabola is symmetric with
respect to the polar axis.
The intersection A of the curve with its axis of symmetry
is called the vertex, and the axis of
symmetry FA the axis, of the parab-
ola. The segment BB' cut off by
the parabola on the perpendicular to
the axis drawn through the focus is
called the latus rectum; its length
is 4 a, if 2 a is the distance between
focus and directrix. Notice also that
the vertex A bisects this distance
FD so that the distance between focus
and vertex as well as that between vertex and directrix is a.
In Fig. 63 the polar axis is taken positive in the sense from
the pole toward the directrix. If the sense from the directrix
to the pole is taken as positive (Fig. 64), we have again with
F as pole FP = r, but the distance of P from the directrix is
2 a-\-r cos </>, so that the polar equation is now
(2) ^=.— ^^-
^ ^ 1 — cos<^
We have assumed a as a positive number, 2 a denoting the
absolute value of the distance between the fixed point (focus)
and the fixed line (directrix). The radius vector r is then
always positive. But the equations (1) and (2) still represent
parabolas if a is a negative number, viz. (1) the parabola of
Fig. 64, (2) the parabola of Fig. 63, the radius vector r being
negative (§ 16),
Fig. G4
IX, § 175]
THE PARABOLA
171
Q
llllllrm;;.^
4
/
D
d
a\aF
Fig. 65
174. Mechanical Construction. A mechanism for tracing
an arc of a parabola consists of a right-
angled triangle (shaded in Fig. 65), one of
whose sides is applied to the directrix.
At a point R of the other side J?Q a
string of length i^Q is attached ; the other
end of the string is attached at the focus
F. As the triangle slides along the di-
rectrix, the string is kept taut by means
of a pencil at P which traces the parabola.
Of course, only a portion of the parabola can thus be traced,
since the curve extends to infinity.
175. Transformation to Cartesian Coordinates. To obtain
the cartesian equation of the parabola let the origin be taken
at the vertex, i.e. midway between the fixed line and fixed
point, and the axis Ox along the axis of the parabola, positive
in the sense from vertex to focus (Fig. Q>Q>). Then the focus
F has the coordinates a, 0, and the equation of the directrix is
X = —a. The distance FP of any point
P{x, y) of the parabola from the focus is
therefore V(a; — ay -f- 2/^ and the dis-
tance QP of P from the directrix is
a + x. Hence the equation is
{x-ay + y^={a^x)\
which reduces at once to
(3) 2/2 = 4 ax. Fig. 66
This then is the cartesian equation of the parabola, referred
to vertex and axis, I.e. when the vertex is taken as origin and
the axis of the parabola (from vertex toward focus) as axis Ox.
Notice that the ordinate at the focus (a, 0) is of length 2 a ;
the double ordinate B'B at the focus is the latus rectum (§ 173).
172
PLANE ANALYTIC GEOMETRY [IX, § 176
176. Negative Values of a. In the last article the constant
a was again regarded as positive ; but (compare § 173) the equa-
tion (3) still represents a parabola when a is a negative number,
the only difference being that in this case the parabola turns its
opening in the negative sense of the axis Ox (toward the left
in Fig. 66). Thus the parabolas y^=4:ax and 2/2= — 4 ax are sym-
metric to each other with respect to the axis Oy (Ex. 14, p. 138).
The equation (3) is very convenient for plotting a parabola
by points. Sketch, with respect to the same axes, the parab-
olas : y^ = 16xj y^ = — 16 x, y^ = x, y^=z — Xj y'^=3xj y'^=z — \ x,
177. Axis Vertical. The equation
(4) x^^^ay,
which differs from (3) merely by the interchange of x and y,
evidently represents a parabola whose vertex lies at the origin
and whose axis coincides with the axis Oy. The parabolas (3)
and (4) are each the reflection of the other in the line y =x
(Ex. 14, p. 138). The equation (4) can be written in the form
1
y
4a
x\
As 1/4 a may be any constant, this is the equation discussed in
§132.
178. New Origin. An equation of the form (Fig. 67)
(5) (2/-/c)2 = 4a(a^-/0,
__Vi^-.
Q
Fig. 68
or of the form (Fig. 68)
(6) (x-hY = 4.a{y-k\
IX, § 179] THE PARABOLA 173
evidently represents a parabola whose vertex is the point (^, k),
while the axis is in the former case parallel to Ox, in the latter
to Oy. For, by taking the point (/i, k) as new origin we can
reduce these equations to the forms (3), (4), respectively.
The parabola (5) turns its opening to the right or left, the
parabola (6) upward or downward, according as 4 a is positive
or negative.
179. General Equation. The equations (5), (6) as well as
the equations (3), (4) are of the second degree. Now the
general equation of the second degree (§ 79),
Ax^ + 2 Hxy -^By^-\-2Gx-{-2Fy-\-C=0,
can be reduced to one of the forms (5), (6) if it contains no
term in xy and only one of the terms in x"^ and y^, i.e. if H =
and either yl or jB is =0. This reduction is performed (as in
§ 80) by completing the square my ov x according as the equa-
tion contains the term in y"^ or x"^.
Thus any equation of the second degree, containing no term in
xy and only one of the squares x^, y"^, represents a parabola, whose
vertex is found by completing the square and whose axis is
parallel to one of the axes of coordinates.
EXERCISES
1. Sketch the following parabolas :
(a) r = ? (6) r = — (c) r = a sec2 4 0.
■^ 1+COS0 ^ ^ 1-COS0
2. Sketch the following curves and find their intersections :
2 CL
(a) r = 8 cos <f>^ r = (6) r = a, r =
1 — cos 1 + cos
8 2 6E
(c) r = 4 cos 0, r = (d) r cos<p = 2 a, r =
1 4- cos ^ ^ 1 - cos
3. Sketch the following parabolas :
(a) (y-2y = S(x^6). (b) (x + Sy = b(S ^ y).
(c) x2 = 6(1/ + 1). (d) (y + 3)2 = - 3 X,
174 PLANE ANALYTIC GEOMETRY [IX, § 179
4. Sketch each of the following parabolas and find the coordinates of
the vertex and focus, and the equations of the directrix and axis :
^ (a) y^-2y-3x-2 = 0. - (6) a;2 + 4 a; - 4 ?/ = 0.
(c) x'^-Ax + Sy + l =0. (d) Sx^-6x- y = 0.
(c) 8y'^-16y-^x + 6 = 0. (/) y2^y + x = 0.
(gr) x2 - X - 3 ?/ + 4 = 0. (/i) 8 ?/2 - 3x + 3 = 0.
6. Sketch the following loci and find their intersections :
' («) y = 2x, y = x2. (&) ?/^ = 4 ax, x + ?/ = 3 a.
(c) y^ = x + S, 2/2 = 6- X. (d) ?/2+4x+4=0, x2 + ?/2^41.
6. Sketch the parabolas with the following line^ and points as direc-
trices and foci, and find their equations :
— (a) X - 4 = 0, (6, - 2). (6) y + 3 = 0, (0, 0).
(c) 2x + 5 = 0, (0, -1). (d) x = 0, (2, -3).
(e) 3y-l=0, (-2, 1). (/) x - 2 a = 0, (a, b).
7. Find the parabola, with axis parallel to Ox, and passing through
the points :
— (a) (1,0), (5,4), (10, -6). (&) (-W, -5), (|, 0), (|, -3).
(c) (-1, 6), (3,1), (-V-,0).
8. Find the parabola, with axis parallel to Ojy, and passing through
the points :
-• (a) (0, 0), (-2, 1), (6, 9). (6) (1, 4), (4, - 1), (-3, 20).
^(c) (-2,1), (2, -7), (-3, -2).
9. Find the parabola whose directrix is the line 3x — 4y— 10 =
and whose focus is: (a) at the origin ; (b) at (5, — 2). Sketch each of
these parabolas. When does the equation of a parabola contain an xy
term ?
10. Find the parabolas with the following points as vertices and foci
(two solutions) :
-(a) (-3, 2), (-3, 5). (6) (2, 5), (- 1, 5).
(c) (- 1, - 1), (1, - 1). {d) (0, 0), (0, - a).
11. Show that the area of a triangle whose vertices Pi (xi, yi),
P2 {X2, 2/2), P3 (X3 , 2/3") are on the parabola 2/2 = 4 ax, may be expressed
by the determinant
1/. 1
= ^(^2 - 2/3) (^3 - yi){y2 -yi)'
o a
1
yi^ yi 1
8a
2/2^ 2/2 I
2/3^ 2/3 1
IX, § 179] THE PARABOLA 175
12. The area J, of a cross-section of a sphere of radius B, at a distance
h from the surface, is given by the formula
A = 2Ilh-h^ h<B.
Reduce this equation to standard form A = kh^, where A and h differ
from A and h by constants. What is the meaning of A and h ?
13. Show that if the area A of the cross-section of any solid perpen-
dicular to a line Z, at a distance h from any fixed point P in Z, is a quad-
ratic function of h :
A = ah'^ + &A. -t- c ;
another point Q in I exists, such that
A = kh^,
where h denotes the distance from Q and A differs from J. by a constant.
14. If s denotes the distance (in feet) from a point P in the line
of motion of a falling body, at a time t (in seconds),
where g is the gravitational constant (32.2 approximately) and Sq is the
distance from P at the time Iq, show that this equation can be put in
the standard form
s = hgT,
where s denotes the distance from some other fixed point in the line of
motion and tis the time since the body was at that point.
15. The melting point t (in degrees Centigrade) of an alloy of lead and
zinc is found to be
t = 13S+ .S16x + .01125 x%
where x is the percentage of lead in the alloy. Reduce the equation to
standard form t = kx \ and show that x —x — U^ t = t — k, where h is
the percentage of lead that gives the lowest melting point, and k is the
temperature at which that alloy melts.
16. Show that the locus of the center of the circle which passes
through a fixed point and is tangent to a fixed line is a parabola.
17. Show that the locus of the center of a circle which is tangent to a
fixed line and a fixed circle is a parabola. Find the directrix of this
parabola.
18. Write in determinant form the equation of the parabola through
three given points, Pi{xx, ^/l), Pi{x2, 1/2), Psi^s, 2/3) with axis parallel
to a coordinate axis.
176
PLANE ANALYTIC GEOMETRY [IX, § 180
180. Slope of the Parabola. The slope tan a of the parabola
2/2 = 4 aa;
at any point P (x, y) (Fig. 69) can be found (comp. § 137) by
first determining the slope
tan «! = y^^y
of the secant PP^ , and then letting
-f*i(^i? Vi) move along the curve up
to the point P(x, y). Now as Pj
comes to coincide with P, x^ becomes
equal to x, and y^ equal to y, so that
the expression for tan a^ loses its
meaning. But observing that P and
Pi lie on the parabola, we have y"^ = 4 ax and y^
hence y^ — y'^ = 4ta(x^ — x). Substituting from this relation
the value of x^ — x in the above expression for tan cti, we find
for the slope of the secant :
4 axy^ , and
tan «i = 4 a -^ — — =
1 o o
4a
2/i -r 2/1 + 2/
If we now let Pj come to coincidence with P so that y^ becomes
= y, we find for the slope of the tangent at P(x, y) :
(7)
, 2a
tana =
y
This slope of the tangent at P is also called the slope of the
parabola at P. The ordinate y of the parabola is a function of
the abscissa x ; and the slope of the parabola at P (x, y) is the
rate at which y increases with increasing £c at P; in other words,
it is the derivative y' of y with respect to x (compare § 138).
As by the equation of the parabola we have y = ± 2^ ax, we
find:
IX, § 182] THE PARABOLA 177
(8) 2/' = tana = ?^=±J«.
y ^x
The double sign in the last expression corresponds to the fact
that to a given value of x belong two points of the curve with
equal and opposite slopes.
I 181. Explicit and Implicit Functions. The result just obtained
that when 2/2 = 4 ax then the derivative of y with respect to x is
y
can he derived more easily by the general method of the differential cal-
culus. This requires, however, some preliminary explanations.
In the cases in which we have previously determined the derivative y'
of a function yotx this function was given explicitly ; i.e. the equation be-
tween X and y that represents the curve was given solved for ?/, in the
form y=f(x).
Our present equation of the parabola, ?/2 = 4 ax, is not solved for y
(though it could readily be solved for y by writing it in the form
y = ± 2y/ax) ; the same is true of the equation of the circle x^ + y^ = a^,
or more generally x^ + y'^ + ax -\- by -{- c = 0, and also of the general equa-
tion of the second degree (§79), Ax^+2 Hxy + By^ +2 Gx + 2 Fy+G =0.
Such equations in x and y, whether they can be solved for y or not, are
said to give y implicitly as a function of x. For, to any particular value
of x we can find from such an equation the corresponding values of x
(there may be several values ; and they may be real or imaginary). Thus,
any equation between x and y, of whatever form, determines y as a func-
tion of X.
182. Derivatives of Implicit Fimctions. The differential cal-
culus shows that to find the derivative ?/' of a function y given implicitly
by an equation between x and y we have only to differentiate this equation
with respect to x, i.e. to find the derivative of each term, remembering
that y is a function of x. To do this in the simple cases with which we
shall have to deal we need only the following two propositions {A) and
(5), §§ 183, 184.
N
178 PLANE ANALYTIC GEOMETRY [IX, § 183
183. (A) Derivative of a Function of a Function, if u is a
function of y, and y a function of x, the derivative of u with respect to x
is the product of the derivative of u with respect to y into the derivative y'
of y with respect to x.
For, as u is a function of y which itself is a function of x, u is also
a function of a;. If x be increased by Ax, y will receive an increment Ay
and u an increment Aw. We want to find the derivative of u with respect
to X, i.e. the limit of Au/Ax as Ax approaches zero. Now we can put
Am _ All Ay .
Ax Ay Ax'
the limit of the first factor, Au/Ay, is the derivative of u with respect to
y, while the limit of the second factor. Ay/ Ax, is the derivative y' of y
with respect to x.
Thus, ii u = y", we know (§ 151) that the derivative of ii with respect
to y is = ny'^~'^. But if u = «/", and ?/ is a function of x, we can also find
the derivative of u loith respect to x; by the proposition {A) it is
ny^~^ ' y' . For example, suppose that ?« = ?/^, where i/=:x2 — .Sx, so
that u = (x2 — 3 x)^. Then the ^/-derivative of u is 3 y'^ ; but the x-de-
rivative of m is 3 ?/2 . y' = 3 y^{^ x - 3) = 3(x2 - 3 x)2(2 x - 3). This can
readily be verified by expanding (x'-^ — 3x)3 and differentiating the result-
ing polynomial in the usual way (§ 150).
184. {B) Derivative of a Product, if n and v are functions of x,
the derivative of uv is u times the derivative of v plus v times the deriva-
tive of u :
derivative of uv = nv' + vu'.
For, putting uv = y, we have to find the limit of Ay / Ax. When x is in-
creased by Ax, u receives an increment Am, v an increment Av, and the
increment Ay of y is therefore
Ay = {u + Au){v + Av) — uv ;
dividing by Ax, we find
^ = {n ^ Au){y ^ Av) - uv ^^A?;_^^Am_^Am^^_
Ax Ax Ax Ax Ax,
In the limit. Ay / Ax becomes «/', Av/Ax becomes u'; Auj A.x becomes u\
and the last term vanishes because its factor At? becomes zero. Hence :
?/' 3= uv' + vuK
IX, § 185] THE PARABOLA 179
*
185. Computation of Derivatives of Lnplicit Functions.
We are now prepared to find the derivative of y when y is given im-
plicitly as a function of x by the equation y'^ = 4 aa;. We have only
to differentiate this equation with respect to a;, i.e. find the x-derivative
of each term, rememhering that y is a function of x. The term y^^ as
a function of a function, gives 2y ■ y' ; the teiin 4 ax gives 4 a ; hence
we find
2 yy' =z 4: a, whence y' = — ,
y
as in § 180.
Similarly, we find by differentiating the equation of the circle
x2 +2/2= a2
that 2x+2yy' =0,
whence y' = — -;
y
i.e. the slope of the circle x^ + y^ = cfi at any point P(a;, y) is minus the
reciprocal of the slope of the radius through P.
If y is given implicitly as a function of x by the equation
x2+ 5x?/= 12,
which, as we shall see later, represents a hyperbola, we find the derivative
of «/, i.e. the slope of the hyperbola, by differentiating the equation and
applying to the second term the proposition {B) :
2x+6x-y' + ?/-5 = 0,
whence y/ ^ _ 5 y + 2 x ^_ y _ 2
5x . X b
EXERCISES
1. Find the derivative of u with respect to x for the following
functions :
— (a) M = 2/2, when 1/ =3x — 5, (6) u = y^-\-'^y.,Yf\iQr).y—x^—2x.
-~{c) M = 2y3— 3?/2,when y=x^-^x. {d) u= ly^ — y, when y = x^.
2. Find the slope of the following parabolas at the point P(x, y) :
_(a)y^ = bx. (&) 2/2_5y + 6x+4 = 0. -(c) 3 2/2 = 4 x- 5.
3. Find y' for the following products :
^(a) y =x\x^ + 6x). (6) y = (x + S)(x- 6).
(c) y = (x-a)(x-b)(x-c). (d) y =^ (x^ S)(2x + 1).
180 PLANE ANALYTIC GEOMETRY [IX, § 185
4. Find the slope at the point P(a-, y) for each of the following circles
by differentiation ; compare the results with §§ 88, 89 :
(a) x2 + y2 = 12. (6) x^ + y^ + ax + by + c = 0.
(c) Ax"^ -h Ay^ -^ 2 Gx -\- 2 Fy + C = 0.
\6. Find -the slope y' for each of the following curves at the point
P(x, y) :
(a) xy = cCK (b) x^y - 6x -\- 4 = 0.
(c) u4x2 + 2 Hxy -{- By^ + 2Gx-\-2Fy+ C = 0.
186. Equation of the Tangent. As the slope of the
parabola f- = 4.ax
at the point P{x,y) is 2a/y (§§180-185), the equation of the
tangent at this point is
F-2/ = — (X-a;),
y
where X, Y are the coordinates of any point of the tangent,
while Xj y are the coordinates of the point of contact. This
equation can be simplified by multiplying both sides by y
and observing that ?/^ = 4 ax ; we thus find
(9) yY=2a{x+X).
•Notice that (as in the case of the circle, § 89) the equation
of the tangent is obtained from the equation of the curve,
7/2 = 4 ax, by replacing y"^ hj yY,2 xhj x-{- X.
The segment TP (Fig. 70) of the tangent from its intersec-
tion T with the axis of the
parabola to the point of contact
P is called the leyigth of the
tangent at P; the projection TQ
of this segment TP on the axis
of the parabola is called the \
subtangent at P. Now, with ^i^- "^^
F=0, equation (9) gives X=—x, i.e. T0= OQ; hence the
subtangent is bisected by the vertex. This furnishes a simple
IX, § 188] THE PARABOLA 181
construction for the tangent at any point P of the parabola if
the axis and vertex of the parabola are known.
187. Equation of the Normal. The normal at a point P
of any plane curve is defined as the perpendicular to the tan-
gent through the point of contact.
The slope of the normal is therefore (§ 27) minus the recip-
rocal of that of the tangent. Hence the equation of the normal
to the parabola is :
r-,=-A(x-.),
that is :
(10) yX-^2aY={2a-\-x)y.
The segment PN of the normal from the point P{x, y)
on the curve to the intersection N of the normal with the axis
of the parabola is called the length of the normal at P; the
projection QN oi this segment P^on the axis of the parabola
is called the subnormal at P.
Now, with Y= 0, equation (10) gives X = 2 a -j- a;, and as
x=OQ, it follows that QN'—2a', i.e. the subnormal of the
parabola is constant, viz. equal to half the latus rectum.
188. Intersections of a Line and a Parabola. The inter-
sections of the parabola
7/2 = 4 aa;
with the straight line
y = mx -f b
are found by substituting the value of y from the latter in the
former equation :
(mx -\-by = 4: ax,
or, reducing:
m V + 2 (m6 - 2 a) i» + &^ = 0.
The roots of this quadratic in x are the abscissas of the
points of intersection ; the ordinates are then found from
y = mx -f- b.
182 PLANE ANALYTIC GEOMETRY [IX, § 188
It thus appears that a straight line cannot intersect a parabola
in more than two points. If the roots are imaginary, the line
does not meet the parabola; if they are real and equal, the
line has but one point in common with the parabola and is
a tangent to the parabola (provided m ^ 0).
189. Slope Equation of the Tangent. The condition for
equal roots is
(bm-2af = b'm\
which reduces to
m = «.
b
The point that the line of this slope has in common with the
parabola is then found to have the coordinates
2 a — bm b^ , , o i.
m^ a
As the slope of the parabola at any point {x, y) is (§ 180)
I/' = 2 a/y, the slope at the point just found is y' = a/b = m ;
i.e. the slope of the parabola is the same as that of the line
y = mx-\-b; this line is therefore a tangent. Thus, the line
(11) y = mx H —
m
is tangent to the parabola y^ = 4: ax whatever the value of m.
This may be called the slope-form of the equation of the tangent.
Equation (11) can also be deduced from the equation (9), by
putting 2 a/y = m and observing that 2/^ = 4 ax.
190. Slope Equation of the Normal. The equation (10) of
the normal can be written in the form
2a 2a
or since by the equation (3) of the parabola x = y^/A a :
Y=-JLx + y + -^'
2a ^^^Sa'
IX, § 191] THE PARABOLA 183
If we denote by n the slope of this normal, we have :
71=-^, y = -2an, J--=-an\
2 a Sa^
so that the equation of the normal assumes the form
(12) r= nX-2a7i- an^.
This may be called the slope-form of the equation of the normal.
191. Tangents from an Exterior Point. The slope-form
(11) of the tangent shows that /rom any j^oint (x, y) of the plane
not more than two tangents can he drawn to the parabola 2/^ = 4 ax.
For, the slopes of these tangents are found by substituting in
(11) for X, y the coordinates of the given point and solving the
resulting quadratic in m. This quadratic may have real and
different, real and equal, or complex roots.
Those points of the plane for which the roots are real and
different are said to lie outside the parabola ; those points for
which the roots are imaginary are said to lie within the parab-
ola; those points for which the roots are equal lie on the
parabola.
The quadratic in m can be written
xm^ — ym -f- a = 0,
so that the discriminant is 2/^ — 4 ax. Therefore a point (x, y)
of the plane lies within, on, or outside the parabola according as
y^ — 4:ax is less than, equal to, or greater than zero.
Similarly, the slope-form (12) of the normal shows that not
more than three normals can be drawn from any point of the
plane to the parabola, since the equation (12) is a cubic for n
when the coordinates of any point of the plane are substituted
for X, Y. As a cubic has always at least one real root (§ 160),
there always exists one normal through a given point; but
there may be two or three.
184
PLANE ANALYTIC GEOMETRY [IX, § 192
192. Geometric Properties. Let the tangent and normal
at P (Fig. 71) meet the axis at T, N; let Q be the foot of the
perpendicular from P to
the axis, D that of the per-
pendicular to the directrix
d ; and let be the vertex,
F the focus.
As the subtangent TQ is
bisected by (§ 186) and
J the subnormal QN is equal
to 2 a (§ 187), while 0F=
a, it follows that F lies
midway between T and TV.
The triangle TPN being Fig. 71
right-angled at P and F being the midpoint of its hypotenuse,
it follows that ^^p _ ^rp_ ^-^
Hence, if axis and focus are given, the tangent and the normal
at any point P of the parabola are found by describing about
F a circle through P which will meet the axis at T and N.
As FP=DP, it follows that FPDT is a rhombus; the
diagonals PT and FD bisect therefore the angles of the
rhombus and intersect at right angles. As TP (like TQ) is
bisected by the tangent at the vertex, the intersection of these
diagonals lies on this tangent at* the vertex. The properties
just proved that the tangent at P bisects the angle between the
focal radius PF and the parallel PD to the axis and that the
perpendicular from the focus to the tangent meets the tangent on
the tangent at the vertex are of particular importance.
193. Diameters. It is known from elementary geometry that
in a circle all chords parallel to any given direction have their
midpoints on a straight line which is a diameter of the circle.
IX, § 193]
THE PARABOLA
185
Similarly, in a parabola, the locus of the midpoiyits of all chords
parallel to any given direction is a straight line, and this line
which is parallel to the axis
is called a diameter of the
parabola. To prove this, take
the vertex as origin and the
axis of the parabola as axis Ox
(Fig. 72) so that the equation
is / = 4 ax. Any line of given
slope m has the equation
y = mx 4- 6,
and with variable b this represents a pencil of parallel lines.
Eliminating x we find for y the quadratic
Fig. 72
r
i^2/ + — = 0.
m m
The roots ^i, 2/2 are the ordinates of the points P,, P^ at
which the line intersects the parabola. The sum of the roots is
4 a
2/1 + ^2 = — ;
m
hence the ordinate \{yi + 2/2) of the midpoint P between P^ , P^
is constant (i.e. independent of x), viz. = 2 a/m, and independ-
ent of b. The midpoints of all chords of the same slope m
lie, therefore, on a parallel to the axis, at the distance 2 a/m
from it.
The condition for equal roots (§ 189) gives b = a/m. That
one of the parallels which passes through the point where the
diameter meets the parabola is, therefore,
, a
y = mx-^—]
m
by § 189 this is a tangent. Thus, the tangent at the end of a
diameter is parallel to the chords bisected by the diameter.
[^
186 PLANE ANALYTIC GEOMETRY [IX, § 193
EXERCISES
1. Find and sketch the tangent and normal of the following parabolas
at the given points :
(a) 2^2 = 25 X, (2, 5). (b) Si/ =4x, {S, - 2). (c) y^ = 2x,{^,l).
(d) 5y^=12x,{l-2). (e) i/^ = x,{hl). (/) 452/2 = x, (5, 1).
2. Show that the secant through the points P(x, y) and Pi (xi , yi)
of the parabola i/2 = 4 ax has the equation 4:aX—(y+yi)Y+yyi = 0,
and that this reduces to the tangent at P when Pi and P coincide.
3. Find the angle between the tangents to a parabola at the vertex
and at the end of the latus rectum. Show that the tangents at the ends of
the latus rectum are at right angles.
4. Find the length of the tangent, subtangent, normal, and subnormal
of the parabola y'^ z=4xa,t the point (1, 2).
5. Find and sketch the tangents to the parabola y'^ = Sx from each
of the following points :
(a) (- 2, 3). (b) (- 2, 0). (c) (- 6, 0). (d) (8, 8).
6. Draw the tangents to the parabola y'^ =3x that are inclined to the
axis Ox at the angles : («) 30°, (6) 45^, (c) 135°, (d) 150° ; and find
their equations.
7. Find and sketch the tangents to the parabola ?/2 = 4 x that pass
through the point (—2, 2).
8. Find and sketch the normals to the parabola y^ = 6x that pass
through the points :
(«) (1,0). (6)(V-, -3). (c)(-V, -f). (^)(f,-|). (e) (0,0).
9. Are the following points inside, outside, or on the parabola
Sy^ = x? (a) (3,1). (5) (2, J), (c) (8, |). (c?) (10, f).
10. Show that any tangent to a parabola intersects the directrix and
latus rectum (produced) in points equally distant from the focus.
11. Show that the tangents drawn to a parabola from any point of the
directrix are perpendicular.
12. Show that the ordinate of the intersection of any two tangents to
the parabola y^ = i ax is the arithmetic mean of the ordlnates of the
points of contact, and the abscissa is the geometric mean of the abscissas
of the points of contact.
IX, § 193] ^ THE PARABOLA 187
13. Show that the sum of the slopes of any two tangents of the parab-
ola y^ = 4 ax is equal to the slope Y/Xof the radius vector of the point of
intersection (X, Y) of the tangents ; find the product of the slopes.
14. Find the locus of the intersection of two tangents to the parabola
2/2 = 4 ax, if the sum of the slopes of the tangents is constant.
-^ 15. Find the locus of the intersection of two perpendicular tangents to
a parabola ; of two perpendicular normals to a parabola.
16. Show that the angle between any two tangents to a parabola is
half the angle between the focal radii of the points of contact. %vCy^\ li**'/^.
17. From the vertex of a parabola any two perpendicular lines are
drawn ; show that the line joining their other intersections with the
parabola cuts the axis at a fixed point.
18. Find and sketch the diameter of the parabola y^ = 6x that bisects
the chords parallel to Sx — 2y-\-5 = 0; give the equation of the focal
chord of this system.
19. Find the system of parallel chords of the parabola y^ = Sx bisected
by the line y = S.
20. Find the diameter and corresponding chord of the parabola y^=^x
/ that pass through the point (5, —2) ; at what angle does this diameter
f* meet its chord ?
21. Show that the tangents at the extremities of any chord of a parab-
ola intersect on the diameter bisecting this chord. Compare Ex. 12.
22. Find the length of the focal chord of a parabola of given slope m.
23. Find the tangent and normal to the parabola x^ = 4 ay in terms of
the coordinates of the point of contact.
24. Find the angles at which the parabolas y^ = i.ax and x^ = 4ay
intersect.
25. If the vertex of a right angle moves along a fixed line while one
side of the angle always passes through a fixed point, the other side
envelopes a parabola (i.e. is always a tangent to the parabola) . The fixed
line is the tangent at the vertex, the fixed point is the focus of the
parabola.
26. Two equal confocal parabolas have the same axis but open in op-
posite sense ; show that they intersect at right angles.
X \rt
188 PLANE ANALYTIC GEOMETRY [IX, § 193
27. If axis, vertex, and one other point of the parabola are given, ad-
ditional points can be constructed as follows : Let O be the vertex, P the
given point, and Q the foot of the perpendicular from P to the tangent
at the vertex ; divide QF into equal parts by the points A\, ^2, ••• ; and
OQ into the same number of equal parts by the points By, P2, ••• ; the
intersections of O^i, OA2, ••• with the parallels to the axis through Pi,
P2, ••• are points of the parabola.
28. If two tangents AP^ AP2 to a parabola with their points of con-
tact Pi, P2 are given and ^Pi, AP2 be divided into the same number of
equal parts, the points of division being numbered from Pi to A and from
A to P2, the lines joining the points bearing equal numbers are tangents
to the parabola. To prove this show that the intersections of any tangent
with the lines ^Pi, ^P2 divide the segments Pi^, J.P2 in the same
division ratio.
29. The shape assumed by a uniform chain or cable suspended between
two fixed points Pi, P2 is called a catenary ; its equation is not algebraic
and cannot be given here. But when the line P1P2 is nearly horizontal
and the depth of the lowest point below P1P2 is small in comparison with
P1P2, the catenary agrees very nearly with a parabola.
The distance between two telegraph poles is 120 ft. ; P2 lies 2 ft. above
the level of Pi ; and the lowest point of the wire is at 1/3 the distance be-
tween the poles. Find the equation of the parabola referred to Pi as
origin and the horizontal line through Pi as axis Ox ; determine the posi-
tion of the lowest point and the ordinates at intervals of 20 ft.
30. The cable of a suspension bridge assumes the shape of a parabola
if the weight of the suspended roadbed (together with that of the cables)
is uniformly distributed horizontally. Suppose the towers of a bridge
240 ft. long are 60 ft. high and the lowest point of the cables is 20 ft. above
the roadway ; find the vertical distances from the roadway to the cables
at intervals of 20 ft.
31. When a parabola revolves about its axis, it generates a surface called
a paraboloid of revolution ; all meridian sections (sections through the
axis) are equal parabolas. If the mirror of a reflecting telescope is such
a surface (the portion about the vertex) , all rays of light falling in parallel
to the axis are reflected to the same point ; explain why.
IX, § 195] THE PARABOLA 189
194. Parameter Equations. Instead of using the cartesian
or polar equation of a curve it is often more convenient to
express x and y (or r and <^) each in terms of a third variable,
which is then called the parameter.
Thus the parameter equations of a circle of radius a about the
origin as center are :
x = a cos </), y = a sin <j>,
<f) being the parameter. To every value of <^ corresponds a
definite x and a definite y, and hence a point of the curve.
The elimination of <f), by squaring and adding the equations,
gives the cartesian equation x^-^y^ = o^.
Again, to determine the motion of a projectile we may observe
that, if gravity were not acting, the projectile, started with an
initial velocity v^ at an angle c to the horizon would have at the
time t the position
a; = Vo cos € • ^, ?/ = -Vo sin c • t,
the horizontal as well as the vertical motion being uniform.
But, owing to the constant acceleration g of gravity (down-
ward), the ordinate y is diminished by ^gt"^ in the time tj so
that the coordinates of the projectile at the time t are
x = Vi) cos c • ^, y — VQ^mc 't — ^ gt\
These are the parameter equations of the path, the parameter
here being the time t. ■ The elimination of t gives the cartesian
equation of the parabola described by the projectile :
y = Vota.n€'X- J^ x\
2 Vq cos^ c
195. Parameter Equations of a Parabola. For any parabola
2/2 = 4 dec we can also use as parameter the angle a made by the
tangent with the axis Ox-, we have for this angle (§ 180) :
, 2a
tana = — ;
y
it follows that y = 2a ctn a and hence x = y'^/A: a= a ctn^ a.
190 PLANE ANALYTIC GEOMETRY [IX, § 195
The equations
X — a ctn^ a, y = 2 a ctn a
are paramenter equations of the parabola y^ = 4:ax; the elimina-
tion of cot a gives the cartesian equation.
196. Parabola referred to Diameter and Tangent. The
equation of the parabola y^ = 4iax preserves this simple form if instead of
axis and tangent at the vertex we take as
axes any diameter and the tangent at its end.
The equation in these oblique coordinates is
yi^ = 4 aixi ,
where ai = a/sin"^ a, a being the angle betvi^een
the axes, i.e. the slope angle at the tiq-w origin
Oi (Fig. 73).
To prove this observe that as the new origin
0\_ {h, k) is a point of the parabola i/2 = 4 ax
we have by § 195
h = a ctn* a, k = 2 a ctn a,
y
/r'
^v
*/
y^
/
/
u
y
/"
h\
/ \
X
/"
/
V
>
Fig. 73
a being the angle at which the tangent at Oi is inclined to the axis.
Hence, transferring to parallel axes through Oi, we obtain the equation
which reduces to
+ 2 a ctn ay = 4 a (x + a ctn^ «),
+ 4 a ctn cc . 1/ = 4 ax.
The relation between the rectangular coordinates x, y and the oblique
coordinates Xi , yi , both with Oi as origin, is seen from the figure to be
X = xi + yx cos a, y = yi sin a.
Substituting these values we find
yi^ sin2 ct + 4 a cos « • ?/i = 4 axi + 4 a!/i cos a.
I.e.
2/1^ = 4
a
sin 2 a
xi = 4 a\X\
if we put a/sin2 a — a\.
IX, § 198]
THE PARABOLA
191
The meaning of the constant ai appears by observing that
sin2 a tan2 ^j
ai =
ai is therefore the distance of the new origin 0\ from the directrix, or
what amounts to the same, from the focus F.
197. Area of Parabolic Segment. A parabola, together with
any chord perpendicular to its axis, bounds an area OPV^ (shaded in
Fig. 74). It was shown by Archimedes (about
250 B.C.) that this area is two thirds the area
of the rectangle PP'Q'Q that has the chord
P'P as one side and the tangent at the vertex
as opposite side. 'Yig. 74
This rectangle PP'Q'Q is often called (somewhat improperly) the cir-
cumscribed rectangle so that the result can be expressed briefly by saying
that the area of the parabola is 2/S of that of the circumscribed rectangle.
This statement is of course equivalent to saying that the (non-shaded)
area OQP is 1/3 of the area of the rectangle OQPB. In this form the
proposition is proved in the next article,
198. Area by Approximation Process. To obtain first an ap-
proximate value {A) for the area OQP (Fig. 75) we may subdivide the
area into rectangular strips of equal width,
by dividing OQ into, say, n equal parts
and drawing the ordinates ?/i , y^, •••?/«.
If the width of these strips is Aic so that
0Q = nAx, we have as approximate value
of the area :
{A) = Aa; . ?/i + Ax . ?/2 +
Fig. 75
+ Ax . yn.
Now yi is the ordinate corresponding to the abscissa Ax ; ?/2 corresponds
to the abscissa 2 Ax, etc. ; ?/„ corresponds to the abscissa wAx = OQ.
Hence, if the equation of the curve is x^ = 4 ay., we have :
?/l=:-L(Ax)2, ?/2 = -1 (2 AX)2,
4a 4a
4a
(wAx)2.
Substituting these values we find :
{A) =
(Ax)3
4a
(1+22 + 32+ ... + W2),
192
PLANE ANALYTIC GEOMETRY [IX, § 198
By Ex. 3 6, p. 74,
1 + 22+ ... +yj2
i «(n + 1)(2 n 4- 1) = 1(2 n3 + 3 n^ + n) ;
hence
(^) = IM'(2n3 + 3w2 + «)
^4 O,
(tiAx)
24
^Y2 +? + !).
a V n n^J
Now nAx = OQ = Xn^ the abscissa of the terminal point P, whatever the
number n and length Ax of the subdivisions. Hence, if we let the num-
ber n increase indefinitely, we find in the limit the exact expression A for
the area OQP:
12a 3 "'4a 3
XnV^
where y„ = Xn^/4 a is the ordinate of the terminal point P. As x^n is
the area of the rectangle OQPE, our proposition is proved.
The integral calculus furnishes a far more simple and more general
method for finding the area under a curve. The method used above
happens to succeed in the simple case of the parabola because we can
express the sum 1 + 2^ + 3^ + ••• + w^ in a simple form.
199. Area expressed in Terms of Ordinates. The area
(shaded in Fig. 76) between the parabola x^ = 4 a?/, the axis Ox, and the
two ordinates 2/1,^3, whose abscissas differ by y
2 Ax is evidently, by the formula of § 198,
^ = _l-(x33-Xi3) = J-[(xi + 2Ax)3-Xin
12 a 12 a
= j^ (6 xi2 + 12 XiAx + 8 (Ax)2).
1^ Gi
Fig. 76
This expression can be given a remarkably
simple form by introducing not only the ordinates y\ — XiV4 a, y% —
(xi + 2 Ax) 2/4 a, but also the ordinate yi midway between yi and 1/3,
whose abscissa is x\ + Ax. For we have :
2/1+4^/2+^3 =i^[^i' + 4(xi + Ax)2 +(xi + 2 Ax)2]
4a
= J_r6xi2 + I2.X1AX + 8(Axj2].
4 a
IX, § 200]
THE PARABOLA
193
y
y.
P^
-^
— ^^
Xt
h
m.
X
Ax Ax
Fia. 77
We find therefore :
^ = |Ax(yi + 4^/2 + ^3).
This formula holds not only when the vertex of the parabola is at the
origin, but also when it is at any point
(A, A;) , provided the axis of the parabola
is parallel to Oy.
For (Fig. 77), to find the area under
the arc F1P2P3 we have only to add to
the doubly shaded area the simply shaded
rectangle whose area is 2 kAx. We find
therefore for the whole area :
\ Ax{yi + 4 «/2 + ys) + 2 A:Aa; = i Ax(yi + 4 ya + 2/3 + 6 fc)
= 1 Aa; [(2/1 + A;) + 4 (^2 + k) +(^3 + A;)],
where yi,y2, 2/3 are the ordinates of the parabola referred to its vertex,
and hence yi -\- k, y2 + k, ys -\- k the ordinates for the origin O.
We have therefore for any parabola whose axis is parallel to Oy :
A = l Ax(yi + 4?/2 + 2/3).
200. Approximation to any Area. Simpson's Rule. The
last formula is sometimes used to find an approximate value for the area
under any curve (i.e. the area bounded
by the axis Ox, an arc AB of the curve,
and the ordinates of A and B, Fig. 78) .
This method is particularly convenient
if a number of equidistant ordinates
of the curve are known, or can be
found graphically.
Let Ax be the distance of the ordi-
nates, and let 2/1,^2, ys be any three
consecutive ordinates. Then the doubly shaded portion of the required
area, between yi and 1/3, will be (if Ax is sufficiently small) very nearly
equal to the area under the parabola that passes through Pi , P2 , P3 and
has its axis parallel to Oy. This parabolic area is by § 199
= ^Aa;(yi+4?/2 +2/3).
The whole area under AB is a sum of such expressions. This method
for finding an approximate expression for the area under any curve is
o
Fig. 78
194
PLANE ANALYTIC GEOMETRY [IX, § 200
known as Simpson's rule (Thomas Simpson, 1743) although the funda-
mental idea of replacing an arc of the curve by a parabolic arc had been
suggested previously by Newton.
Qj Ax Q; Ax Qj
Fig. 79
201. Area of any Parabolic Segment. As the equation of a
parabola referred to any diameter and the tangent at its end has exactly
the same form as when the parabola is referred to its axis and the tan-
gent at the vertex (§ 196) it can easily be shown that the area of any
parabolic segment is 2/3 of the area of the
circumscribed parallelogram. In this
statement the parabolic segment is under-
stood t0j.be bounded by any arc of the
parabola and its chord; and the circum-
scribed parallelogram is meant to have for
two of its sides the chord and the parallel
tangent while the other two sides are
parallels to the axis through the extremities of the chord (Fig. 79).
With the aid of this proposition Simpson's rule can be proved very
simply. For, the area of the parabolic segment P1P3P2 (Fig. 79) is then
equal to 2/3 of the parallelogram formed by the chord P1P2, the tangent
at P2, and the ordinates yi, ys (produced if necessary). This parallelo-
gram has a height = 2 Ax and a base = MP-z = ?/2 — i (2/1 + y^) ; hence
the area of P1P3P2 is
= § Aa; (2 2/2 - yi - 2/3) = i Ar [4 ?/2 - 2 (yi + 2/3)].
To find the whole shaded area we have only to add to this the area of
the trapezoid ^i§3P3Pi which is
= Ax(yi-\-y3).
Hence A = QiQsPsP^Pi = i Ax[4 y,
= I- Ax(yi -f 4 ^2 + yz) .
2(2/1 + 2/3) +3(?/i 4-2/3)]
EXERCISES
1. Show that the area of any parabolic segment is 2/3 of the area
of the circumscribed parallelogram.
2. In what ratio does the parabola y'^ = 4ax divide the area of the
circle (x — a)'^ + y^ = 4:a^?
IX, §201] THE PARABOLA 195
3. Find the area bounded by the parabola i/^ = 4 ax and a line of
slope m through the focus.
4. By a method similar to that used in finding the area of a parabola
(§ 198), find exactly the area bounded by the curve y = 0(fi, the axis Ox,
and the line x = a. Wliat is the area bounded by this same curve, the
axis Ox, and the lines x = a, x = b? What is the area bounded by the
curve y = x^ + c, the axis Ox, and the lines x = a, x = 6 ?
5. Find and sketch the curve whose ordinates represent the area
bounded by : (a) the line ?/ = | x, the axis Ox, and any ordinate, (&) the
parabola y = ^ x^, the axis Ox, and any ordinate.
6. Let Pi(xi, yi), P2(xi + Ax, 2/2), P3(xi + 2 Ax, ys) be three points of
a curve. Let A denote the sum of the areas of the two trapezoids formed
by the chords P1P2 , P2P3 , the axis Ox, and the ordinates yi, y^, ys- Let
B denote the area of the trapezoid formed by any line through P2, the
axis Ox, and the segments cut off on the ordinates yi, ys. Find the
approximation to the area under the curve given by each of the following
formulas: ^(iA + B), 1(2 A + B), l(A + 2B). Which of these gives
Simpson's rule ?
7. To find an approximation to the area bounded by a curve, the axis
Ox, and two ordinates, divide the interval into any even number of strips
of equal width and apply Simpson's rule to each successive pair. Show
that the result found is : the sum of the extreme ordinates plus twice the
sum of the other odd ordinates plus four times the sum of the even ordi-
nates, multiplied by one third the distance between the ordinates.
8. Find an approximation to the areas bounded by the following
curves and the axis Ox (divide the interval in each case into eight or
more equal parts) :
(a) 4y = 16- x\ (6) ?/ = (x + 3) (x - 2)2. (c) y=x'^- x*.
9. "The cross-sections in square feet of a log at intervals of 6 ft. are
3.25, 4.27, 5.34, 6.02, 6.83 ; find the volume.
10. The cross sections of a vessel in square feet measured at intervals
of 3 ft. are 0, 2250, 6800, 8000, 10200 ; find the volume. Allowing one
ton for each 35 cu. ft., what is the displacement of the vessel ?
11. The half-widths in feet of a launch's deck at intervals of 5 ft. are
0, 1.8, 2.6, 3.2, 3.3, 3.3, 2.7, 2.1, 1 ; find the area.
196
PLANE ANALYTIC GEOMETRY [IX, § 202
202. Shearing Force and Bending Moment. A straight
beam AB (Fig. 80), of length I, fixed at one end ^ in a horizontal posi-
tion and loaded uniformlj' with w lb. per unit of length, will bend under
the load. At any point P, at the distance x from A, the efEect of the
load to(Z — x) that rests on PB is ^ ^
twofold :
(a) If the beam were cut at P, yy^
this load, which is equivalent to a ^^
single force W = w{l — x) applied 'f^:A P B
at the midpoint of PP, would pull '■'^ Fig. 80 W-wfl-x)
the portion PB vertically down.
This force which tends to shear off the beam at P is called the shearing
force F at P. Adopting the convention that downward forces are to be
regarded as positive, we have
F=w(l-x).
The shearing force at the various
points of AB is therefore repre-
sented by the ordinates of the
straight line CB (Fig. 81) .
(6) If the beam were hinged at P, the effect of the load 110(1 — x) on PB
would be to turn it about P. As the force w{l — x) can be regarded as
applied at the midpoint of PP, this effect at P is represented by the
bending moment m = -\w{1- xy,
the minus sign arising from the convention of regarding a moment as
positive when tending to turn counterclockwise. As w{l — x) turns
clockwise about P, the moment is ^,^
negative. The curve DB repre-
senting the bending moments
(Fig. 82) is a parabola.
More briefly we may say that
the single force F=w{l — x)
applied at the midpoint of PB
is equivalent to an equal force
c
wl
P
""""■^^-1 '
-A
Fig. 81
^B
Fig. 82
at P, the shear F=w{l--x), together with the couple formed by -f-P
at the midpoint of PB and — P at P ; the moment of this couple is the
bending moment M = — \w{l — xy.
IX, §2031 THE PARABOLA 197
203. Relation of Bending Moment to Shearing Force. For
any beam AB, fixed at one or both ends or supported freely at two or
more points, in a horizontal position, and loaded by any vertical forces,
the shearing force at any point P is defined as the algebraic sum of all the
forces (including the reactions of the supports) on one side of P, and the
bending moment at P as the algebraic sum of the moments of these forces
about P.
It may be noted that if the shear Pis constant, the bending moment is
a linear function of x (i.e. of the abscissa of P) ; if P (as in § 202) is a
linear function of x, M is a. quadratic function ; in either case the deriva-
tive of M with respect to x is equal to P :
M' = F.
It follows that the bending moment is a maximum or minimum at any
point where the shear is zero.
EXERCISES
Determine P and M as functions of x for a horizontal beam AB of
length I and represent Pand ilf graphically :
1. "When the beam is fixed at one end A (cantilever) and carries
a single load W at the other end B.
2. When the beam is freely supported at its ends A, B and loaded :
(a) uniformly with w lb. per unit of length ; (&) with a single load W at
the midpoint ; (c) with a single load W at the distance a from A. De-
termine first the reactions at A and B.
3. When the beam is supported at the two points trisecting it and
carries : (a) a uniform load w lb. /ft. ; (&) a single load W a.t A and at B.
4. When the beam is supported at its ends and is loaded: (a) with
w lb. /ft. over the middle third ; (6) with w lb. /ft. over the first and third
thirds; (c) with w Ib./ft. over the first half and 2 to lb. /ft. over the
second half.
5. When the beam is fixed at A and carries w lb. /ft. over the outer
half.
CHAPTER X
Br-
.>^
FiAt
ELLIPSE AND HYPERBOLA
204. Definition of the Ellipse. The ellipse may be defined
as the locus of a point whose distances from two fixed points have
a constant sum.
If F^ , F2 (Fig. 83 are the fixed points, which are called the
foci, and if P is any point of the
ellipse, the condition to be satisfied ^"
by P is
F^P + F.P = 2 a.
The ellipse can be traced mechan-
ically by attaching at F^, F^ the
ends of a string of length 2 a . and Fig. 83
keeping the string taut by means of a pencil. It is obvious
that the curve will be symmetric with respect to the line i^ii^2>
and also with respect to the perpendicular bisector of F1F2.
These axes of symmetry are called the axes of the ellipse ; their
intersection is called the center of the ellipse.
205. Axes. The points A^, A2, B„ B2 (Figs. 83 and 84)
vrhere the ellipse intersects these axes are called vertices.
The distance ^2 A of those vertices
that lie on the axis containing the
foci Fi, F2 is = 2 a, the length of
the string. For when the point P
in describing the ellipse arrives at
Ai, the string is doubled along
Fi Ai so that Fig. m
198
X, §2061 ELLIPSE AND HYPERBOLA 199
and since, by symmetry, A>F.2 =^ F^A^, we have
^2^2 + F^F^ + i^i A = A A = 2 a.
The distance A2A1 = 2 a, which is called the major axis, must
evidently be not less than the distance F2F1 between the foci,
which we shall denote by 2 c.
The distance B2B1 of the other two vertices is called the
minor axis and will be denoted by 2 b. We then have
for when P arrives at Bi, we have B^F^ = BiFi= a.
206. Equation of the Ellipse. If we take the center as
origin and the axis containing the foci as axis Ox, the equation of
the ellipse is readily found from the condition FiP-\-F2P= 2 a,
which gives, since the coordinates of the foci are c, and
- c, :
^/{x - cf + 2/' 4- V(i^- + c)2 + ?y2 = 2 a.
Squaring both members we have
a;24.2/2_|_c2 4. V(aj2 4.^2_^c'_2 ex) {x'^-\-y'^^c^+2 ex) = 2 a^;
transferring x'^-\-y'^-\-G^ to the right-hand member and squaring
again, we find
i.e. (a2-c2) a;2 4.ay=aXa2-c2).
Now for the ellipse (§ 205) a^-c^=h\ Hence, dividing both
members by aW, we find
as the cartesian equation of the ellipse referred to its axes.
This equation shows at a glance : (a) that the curve is sym-
metric to Ox as well as to Oy ; (b) that the intercepts on the
axes Ox, Oy are ±a, and ±b. The lengths a, b are called the
semi-axes.
200 PLANE ANALYTIC GEOMETRY [X, § 206
Solving the equation for y we find
h
(2) 2/ = ±-Va2-ar',
a
which shows that the curve does not extend beyond the vertex
A^ on the right, nor beyond A2 on the left.
If a and h (or, what amounts to the same, a and c) are given
numerically, we can calculate from (2) the ordi nates of as
many points as we please. If, in particular, a — h (and hence
c = 0) the ellipse reduces to a circle.
EXERCISES
1. Sketch the ellipse of semi-axes a = 4, 6 = 3, by marking the ver-
tices, constructing the foci, and determining a few points of the curve
from the property FiP + F2P = 2 a. Write down the equation of this
ellipse, referred to its axes.
2. Sketch the ellipse x'^/W + y^/9 = 1 by drawing the circumscribed
rectangle and finding some points from the equation solved for y.
3. Sketch the ellipses : (a) x^+2y'^ = l. (6) Sx^-hl2y^ = 5.
(c) 8 a:2 -}- 3 y2 = 20. (d) x^ -^ 20 y^ = 1.
4. If in equation (1) a < ft, the equation represents an ellipse whose
foci lie on Oy. Sketch the ellipses :
(a) ^-1-1^=: 1. (6) 20 x2 -H ?/2 = 1. (c) 10 x2 -H 9 2/2 = 10.
4 16
5. Find the equation of the ellipse referred to its axes when the foci
are midpoints between the center and vertices.
6. Find the product of the slopes of chords joining any point of an
ellipse to the ends of the major axis. What value does this product
assume when the ellipse becomes a circle ?
7. Derive the equation of the ellipse with foci at (0, c), (0, -c), and
major axis 2 a.
8. Write the equations of the following ellipses : (a) with vertices
at (5, 0), (- 5, 0), (0, 4), (0, - 4) ; (b) with foci at (2, 0), (- 2. 0),
and major axis 6.
9. Find the equation of the ellipse with foci at(l, 1), (—1, — 1),
and major axis 6, and sketch the curve.
X, § 208]
ELLIPSE AND HYPERBOLA
201
207. Definition of the Hyperbola. The hyperbola can be
defined as tJie locus of a point whose distances from two fixed
points have a coyistant difference.
The fixed points F^, F^ are again called the foci; if 2 a is
the constant difference, every point P of the hyperbola must
satisfy the condition
F^P-FJ'=±2a.
Notice that the length 2 a must here be not greater than the
distance F^F^ = 2 c of the foci.
The curve is symmetric to the line FiF^ and to its perpen-
dicular bisector.
A mechanism for tracing an arc of a hyperbola consists of
a straightedge F^Q (Fig. 85) which turns about one of the
foci, F2 ; a string, of length F2Q — 2a, is fastened to the
">
Fig. 85
straightedge at Q and with its other end to the other focus,
Fi. As the straightedge turns about F2, the string is kept
taut by means of a pencil at P which describes the hyperbolic
arc. Of course only a portion of the hyperbola can be traced
in this manner.
208. Equation of the Hyperbola. If the line F2F1 be taken
as the axis Ox, its perpendicular bisector as the axis Oy, and if
F2F1 = 2 c, the condition F^P- F^P= ± 2 a becomes (Fig. 86) :
V(x-\-cy-\-f-V(x^cy-hy'=±2a,
202 PLANE ANALYTIC GEOMETRY
Squaring both members we find
[X, § 208
squaring again and reducing as in § 206, we find exactly the
same equation as in § 206 :
Fig. 86
But in the present case c ^ a, while for the ellipse we had
c < a. We put, therefore, for the hyperbola
the equation then reduces to the form
which is the cartesian equation of the hyperbola referred to its axes.
209. Properties of the Hjrperbola. The equation (3) shows
at once: (a) that the curve is symmetric to Ox and to Oy;
(b) that the intercepts on the axis Ox are ± a, and that the
curve does not intersect the axis Oy.
The line F2F1 joining the foci and the perpendicular bisector
of F2F1 are called the axes of the hyperbola ; the intersection
of these axes of symmetry is called the center.
The hyperbola has only two vertices, viz. the intersections
Ai , A2 with the axis containing the foci.
X, §210] ELLIPSE AND HYPERBOLA 203
The shape of the hyperbola is quite different from that of
the ellipse. Solving the equation for y we have
(4) 2/=±-Va^-a^
which shows that the curve extends to infinity from A^^ to the
right and from A^ to the left, but has no real points between
the lines x = a, x = — a.
The line F2F1 containing the foci is called the transverse
axis; the perpendicular bisector of F2F1 is called the conjugate
axis. The lengths a, h are called the transverse and conjugate
semi-cfiXes.
In the particular case when a=b, the equation (3) reduces to
a? — y^ = a^,
and such a hyperbola is called rectangular or equilateral
210. Asymptotes. In sketching the hyperbola (3) or (4) it
is best to draw first of all the two straight lines
i.e.
(5) 2/=±^^,
which are called the asymptotes of the hyperbola.
Comparing with equation (4) it appears that, for any value
of X, the ordinates of the hyperbola (4) are always (in absolute
value) less than those of the lines (5) ; but the difference
becomes less as x increases, approaching zero as x increases in-
definitely.
Thus, the hyperbola approaches its asymptotes more and
more closely, the farther we recede from the center on either
side, without ever reaching these lines at any finite distance
from the center.
204 PLANE ANALYTIC GEOMETRY [X, § 210
EXERCISES
1. Sketch the hyperbola x'^/XQ — y-2/4 = 1, after drawing the asymp-
totes, by determining a few points from the equation solved for y ; mark
the foci.
2. Sketch the rectangular hyperbola cc^ — 2/2 — 9, Why the name
rectangular ?
3. With respect to the same axes draw the hyperbolas :
(a) 20x2 _ 2/2 = 12. (6) a;2 - 20 2/2 = 12. (c) x^ - y"^ = 12.
4. The equation — x2/a2 + y'^/h'^ = 1 represents a hyperbola whose
foci lie on the axis Oy. Sketch the curves :
(a) -3x2 + 42/2 = 24. (^b) x^- Sy^ + 1S = 0. (c) ^2 - 2/^ + 16 = 0.
6. Sketch to the same axes the hyperbolas :
^_y2=l ^_2/2=_i.
9 ^ ' 9 ^
Two such hyperbolas having the same asymptotes are called conjugate.
6. What happens to the hyperbola a;2/a2 _ 2/2/52 = 1 as a varies ? as
b varies ?
7. The equation a;2/a2 — y^/b^ = k represents a family of similar
hyperbolas in which k is the parameter. What happens as k changes
from 1 to — 1 ? What members of this family are conjugate ?
8. Find the foci of the hyperbolas :
(a) 9 x2 - 16 ^2 = 144. (5) 3 a;2 _ y2 = 12.
9. Find the hyperbola with foci (0, 3), (0, — 3) and transvei-se axis 4.
10. Find the equation of the hyperbola referred to its axes when the
distance between the vertices is one half the distance between the foci.
11. Find the distance from an asymptote to a focus of a hyperbola.
12. Show that the product of the distances from any point of a hyper-
bola to its asymptotes is constant. .
13. Find the hyperbola through the point (1, 1) with asymptotes
y = ±2x.
14. Find the equation of the hyperbola whose foci are (1, 1),
(—1, — 1), and transverse axis 2, and sketch the curve.
X, § 212] ELLIPSE AND HYPERBOLA 205
211. Ellipse as Projection of Circle. If a circle be turned
about a diameter A2Ai = 2a through an angle c(<|-7r) and
then projected on the original plane, the projection is an
ellipse.
For, if in the original plane we take the center as origin
and OAi as axis Ox (Fig. 87), the
ordinate QP of every point P of
the projection is the projection of
the corresponding ordinate QP^ of
the circle; i.e.
QP = QPi cos £. Fio. 87
The equation of the projection is therefore obtained from the
equation
' x'^-\- 1/^ = 0^
of the circle by replacing y by y/cos c. The resulting equation
COS^c
represents an ellipse whose semi-axes are a, the radius of the
circle, and b — a cos e, the projection of this radius.
212. Construction of Ellipse from Circle. We have just
seen that, if a > &, the ellipse
a' ¥
can be obtained from its circumscribed circle x^ + y^ = a'^hj re-
ducing all the ordinates of this circle in the ratio b/a. This
also appears by comparing the ordinates
y = ±Wa'-x'
a
of the ellipse with the ordinates y = ± Va^ — x"^ of the circle.
206
PLANE ANALYTIC GEOMETRY [X, § 213
But the same ellipse can also be obtained from its inscribed
circle x^-\-y'^= W by increasing each abscissa in the ratio a/h,
as appears at once by solving for x.
It follows that when the semi-axes a, h are given, points of
the ellipse can be constructed by drawing concentric circles of
radii a, h and a pair of perpendicular diameters (Fig. 88) ; if
y
any radius meets the circles at P^, P^ ? the intersection P of
the parallels through P^ , P^, to the diameters is a point of the
ellipse.
213. Tangent to Ellipse. It follows from § 211 that if
P (x, y) is any point of the ellipse and P^ that point of the cir-
cumscribed circle which has the same abscissa, the tangents at
P to the ellipse and at P^ to the circle must meet at a point T on
the major axis (Fig. 89).
For, as the circle is turned about A^Ai into the position in
which P is the projection of Pj , the tangent to the circle at Pj
is turned into the position whose projection is PT, the point T
on the axis remaining fixed.
X, § 214]
ELLIPSE AND HYPERBOLA
207
The tangent XiX + yiY= o? to the circle at P^ [x^ , 2/1) meets
the axis Ox at the point T whose abscissa is
Hence the equation of the tangent 2XP{x, y) to the ellipse is
X Y 1
x y 1
1
0,
t.e.
yX-fx--]Y-a^^ = 0;
\ xj X
dividing by a^y/x and observing that, by the equation of the
ellipse, a;2 — a^ = — (a'^/b'^)y'^ we find
(6)
a2 ^ 52
as equation of the tangent to the ellipse
a"" b^
at the point P(x, y).
-214. Slope of Ellipse. It follows from the equation of the
tangent that the slope of the ellipse at any point P{x, y) is
¥x
tan a = —
a'y
The slope being the derivative y' can be found more directly by differ-
entiating the equation (1) of the ellipse (remembering that y is a function
of X, compare §§ 181-185) ; this gives
whence
2^ + 2^1^ = 0,
a2 62
?/' = tan «=- — -.
The equation (6) of the tangent is readily derived from this value of
the slope.
208
PLANE ANALYTIC GEOMETRY [X, § 215
215. Eccentricity. For the length of the focal radius F^P
of any point P(x,y) of the ellipse (1) we have (Fig. 90),
since a^ — 6^ = c^ :
Fj^=(x-cy-{-y^={x-cy-\--^{a''-x')=\(a'-- 2 a'^cx-^d'x''),
whence
F,P=±
a x\
a J
The ratio c/a of the distance 2 c of the foci to the major
axis 2 a is called the (numerical)
eccentricity of the ellipse. De-
noting it by e we have
FiP=±(a — ex)j
and similarly we find
F,P=±(a + ex).
For the hyperbola (3) we find in the same way, if we again
put e = c/a, exactly the same expressions for the focal radii
F^P, F2P(m absolute value). Bat as for the ellipse c^^a"^— ¥
while for the hyperbola c^ = a^-{-¥ it follows that the eccentrio-
ity of the ellipse is always a proper fraction becoming zero only
for a circle, while the eccentricity of the hyperbola is always greater
than one. . V
216. Equation of Normal to Ellipse. As the normal to a
curve is the perpendicular to its tangent through the point of
contact, the equation of the normal to the ellipse (1) at the point
P{x, y) is readily found from the equation (6) of the tangent as
lX-^T=xy(^-^\ = ^
¥ a" \b^ ay aW
xy,
I.e.
«'x-^r=c^
X, § 217]
ELLIPSE AND HYPERBOLA
209
Tlie intercept made by this normal on the axis Ox is there-
fore
ON=—x = e'^x.
From this result it appears by § 215 that (Fig. 91)
F^N= c + e^ic = e(a -\-ex)=ze- F,P,
F^N= c - e^a; = e(a -ex)=e' F^P-,
hence the normal divides the dis-
tance F^Fi in the ratio of the
adjacent sides F2P, F^P of the
triangle F.PF^. It follows that
the normal bisects the angle between
the focal radii PFi , PF^ ; in other words, the focal radii are
equally inclined to the tangent.
217. Construction of any H3rperbola from Rectangular
Hyperbola. The ordinates (4),
Fig. 91
y = ±--yx^—a\
a
of the hyperbola (3) are b/a times the corresponding ordinates
y = ± Va^ — a^
of the equilateral hyperbola (end of § 209) having the same
transverse axis. When 6 < a, we can put b/a = cos € and re-
gard the general hyperbola as the projection of the equilateral
hyperbola of equal transverse axis. When 6 > a, we can put
a/b = cos c so that the equilateral hyperbola can be regarded as
the projection of the general hyperbola.
In either case it is clear that the tangents to the general and
equilateral hyperbolas at corresponding points (i.e. at points
having the same abscissa) must intersect on the axis Ox.
210 PLANE ANALYTIC GEOMETRY [X, § 218
218. Slope of Equilateral Hyperbola. To find the slope of
the equilateral hyperbola
x'2 - y^ = a\
observe that the slope of any secant joining the point P(x,y)
and Fi{xi, y^) is {y^ — y)/{x^—x), and that the relations
y''=x^-a?,
yi^ = Xi^-a^
give f- - y,^ = X'' - x,\ i.e. (y - y,)(y + y,) =(x-x,)(x + x^),
whence l^Uh^xJ^^
x-xi y + yi
Hence, in the limit when P^ comes to coincidence with P, we
find for the slope of the tangent at P(x, y) :
tan a = ~'
y
The equation of the tangent to the equilateral hyperbola is
therefore
y
i.e. since x^ —y'^ = a?:
xX-yY=a\
219. Tangent to the Hyperbola. It follows as in § 213 that
the tangent to the geyieral hyperbola (3) has the equation
(7) ^-^=1.
The slope of the hyperbola (3) is therefore
y^x
tan a =
a^y
This slope might of course have been obtained directly by differen-
tiating the equation (3) (compare § 214).
X, §219] ELLIPSE AND HYPERBOLA 211
Notice that the equations (6), (7) of the tangents are obtained
from the equations (1), (3) of the curves by replacing aj^, ip- by
xX^ yY, respectively (compare §§ 89, 186).
It is readily shown (compare § 216) that for the hyperbola
(3) the tangent meets the axis Ox at the point T that divides
the distance of the foci F^F^ proportionally to the focal radii
F^P, FiP, so that the tangent to the hyperbola bisects the angle
between the focal radii.
EXERCISES \ U
1. Show that a right cylinder whose cross-section (i.e. section at
right angles to the generators) is an ellipse of semi-axes a, b has two
(oblique) circular sections of radius a ; find their inclinations to the
cross-section.
2. Derive the equation of the normal to the hyperbola (3) .
3. Find the polar equations of the ellipse and hyperbola, with the
center as pole and the major (transverse) axis as polar axis.
4. Find the lengths of the tangent, subtangent, normal, and sub-
normal in terms of the coordinates at any point of the ellipse.
5. Show that an ellipse and hyperbola with common foci are
orthogonal.
6. Show that the eccentricity of a hyperbola is equal to the secant
of half the angle between the asymptotes.
7. Express the cosine of the angle between the asymptotes of a
hyperbola in terms of its eccentricity.
8. Show that the tangents at the vertices of a hyperbola intersect the
asymptotes at points on the circle about the center through the foci.
9. Show that the point of contact of a tangent to a hyperbola is the
midpoint between its intersections with the asymptotes.
10. Show that the area of the triangle formed by the asymptotes and
any tangent to a hyperbola is constant.
11. Show that the product of the distances from the center of a hyper-
bola to the intersections of any tangent with the asymptotes is constant.
12. Show that the tangent to a hyperbola at any point bisects the angle
between the focal radii of the point. [>^'l^ 4 tUy i^w^X-r^
/^ Z^^ ,, jJHi^
212 PLANE ANALYTIC GEOMETRY [X, § 219
13. As the sum of the focal radii of every point of an ellipse is con-
stant (§ 204) and the normal bisects the angle between the focal radii
(§ 216), a sound wave issuing from one focus is reflected by the ellipse
to the other focus. This is the explanation of " whispering galleries."
Find the semi-axes of an elliptic gallery in which sound is reflected from
one focus to the other at a distance of 69 ft. in 1/10 sec. (the velocity of
/ sound is 1090 ft. /sec).
14. Show that the distance from any point of an equilateral hyperbola
to its center is a mean proportional to the focal radii of the point.
15. Show that the bisector of the angle formed by joining any point
of an equilateral hyperbola to its vertices is parallel to an asymptote.
16. For the ellipse obtained by turning a circle of radius a about a
diameter through an angle e and projecting it on the plane of the circle,
show that the distance between the foci is = 2 a sin e ; in particular,
show that the foci of a circle are at the center.
17. Show that the tangents at the extremities of any diameter (chord
through the center) of an ellipse or hyperbola are parallel.
18. Let the normal at any point Pof an ellipse referred to its axes cut
the coordinate axes at Q and B ; find the ratio PQ/PB.
19. Show that a tangent at any point of the circle circumscribed about
an ellipse is also a tangent to the circle with center at a focus and radius
equal to the focal radius of the corresponding point of the ellipse.
20. Show that the lines joining any point of an ellipse to the ends of
the minor axis intersect the major axis (produced) in points inverse with
respect to the circumscribed circle.
21. Show that the product of the ^/-intercept of the tangent at any
point of an ellipse and the ordinate of the point of contact is constant.
22. Show that the normals to an ellipse through its intersections with
a circle determined by a given point of the minor axis and the foci pass
through the given point.
23. Find the locus of the center of a circle which touches two fixed
non-intersecting circles.
24. Find the locus of a point at which two sounds emitted at an inter-
val of one second at two points 2000 ft. apart are heard simultaneously.
X, § 222] ELLIPSE AND HYPERBOLA 213
220. Intersections of a Straight Line and an Ellipse.
The intersections of the ellipse (1) with any straight line are
found by solving the simultaneous equations
y = mx -\- k.
Eliminating y, we find a quadratic equation in x :
{w?a^ + lf)x^ + 2 mka?x + {k^ - h'^)a' == 0.
To each of the two roots the corresponding value of y results
from the equation y = mx + k.
Thus, a straight line can intersect an ellipse in not more than
two points.
221. Slope Form of Tangent Equations. If the roots of
the quadratic equation are equal, the line has but one point in
common with the ellipse and is a tangent.
The condition for equal roots is
m'^k'^a^ = (m^a"" + b''){k'' - b%
whence k = ± Vm'^a^ -\- ¥.
The two parallel lines
(8) y = mx± Vm^a^ + 6^
are therefore tangents to the ellipse (1), whatever the value of
m. This equation is called the slope form of the equation of a
tangent to the ellipse.
It can be shown in the same way that a straight line cannot
intersect a hyperbola in more than two points, and that the
two parallel lines
y = mx ± Vm^a^ — b^
have each but one point in common with the hyperbola (3).
222. The condition that a line be a tangent to an ellipse or
hyperbola assumes a simple form also when the line is given
in the general form
Ax-hBy-\-C=0.
214 PLANE ANALYTIC GEOMETRY [X, § 222
Substituting the value of y obtained from this equation in
the equation (1) of the ellipse, we find for the abscissas of the
points of intersection the quadratic equation :
{A^o? + B^W)x' H- 2 ACa^x + (C^ -B'¥)a^ = 0;
the condition for equal roots is
which reduces to
The line is therefore a tangent whenever this condition is
satisfied.
When the line is given in the normal form,
X cos p-\-ysm p = p,
the condition becomes
p2 = a2cos2;8-h62sin2^.
223. Tangents from an Exterior Point. By § 221 the line
y = mx + y/m'^a^ + b^
is tangent to the ellipse (1) whatever the value of m. The condition that
this hne pass through any given point (xi , yi) is
yi = mxi + Vm^a^ + b^ ;
transposing the term mxi, and squaring, we find the following quadratic
equation for m :
to2xi2 - 2 mxiyi + yi^ = mH^ + 6^
I.e.* W - a^)w*^ - 2 ^i^iwi + y^ - &2 = 0.
The roots of this equation are the slopes of those lines through ix\ , y{)
that are tangent to the ellipse (I).
Thus, not more than two tangents can be drawn to an ellipse from any
point. Moreover, these tangents are real and different, real and coin-
cident, or imaginary, according as
X, § 225] ELLIPSE AND HYPERBOLA 215
This condition can also be written in the form
6%i2 + a^y{^ = a2&2,
I.e.
Xi'
Hence, to see whether real tangents can be drawn from a point (xi , yi)
to the ellipse (1) we have only to substitute the coordinates of the point
for X, y in the expression
if the expression is zero, the point (xi, yi) lies on the ellipse, and only
one tangent is possible ; if the expression is positive, two real tangents
can be drawn, and the point is said to lie outside the elHpse ; if the expres-
sion is negative, no real tangents exist, and the point is said to lie within
the ellipse.
These definitions of inside and outside agree with what we would
naturally call the inside or outside of the ellipse. But the whole discus-
sion applies equally to the hyperbola (3) where the distinction between
inside and outside is not so obvious.
224. Symmetry. Since the ellipse, as well as the hyperbola,
has two rectangular axes of symmetry, the axes of the curve,
it has a center, the intersection of these axes, i.e. sl point of
symmetry such that every chord through this point is bisected
at this point (compare § 135). Analytically this means that
since the equation (1), as well as (3), is not changed by replac-
ing a; by — x, nor by replacing yhj—y, it is not changed by
replacing both x and y by — x and — ?/, respectively. In other
words, if {x, y) is a point of the curve, so is (— a?, — y). This
fact is expressed by saying that the origin is a point of sym-
metry, or center.
225. Conjugate Diameters. Any chord through the center
of an ellipse or hyperbola is called a diameter of the curve.
216
PLANE ANALYTIC GEOMETRY [X, § 225
Just as in the case of the circle, so for the ellipse the locus
of the midpoints of any system of parallel chords is a diameter.
This follows from the corresponding property of the circle
because the ellipse can be regarded as the projection of a
circle (§211). But this diameter is in general not perpen-
dicular to the parallel chords ; it is said to be conjugate to the
diameter that occurs among the parallel chords. Thus, in Fig.
92, P'Q' is conjugate to PQ (and vice versa).
Fig. 92
To find the diameter conjugate to a given diameter y = mx.
of the ellipse (1), let y=mx-\-khe any parallel to the given
diameter. If this parallel intersects the ellipse (1) at the real
points (flJi, ?/i) and (ajg, 2/2)? t^ie midpoint has the coordinates
^(xi + X2), i(2/i + 2/2)- The quadratic equation of § 220 gives
1 , , V ma^k
X = — (X-, -\- Xo) = ^7 •
If instead of eliminating y we eliminate x, we obtain the quad-
ratic equation
(m^a^.+b^)y^ - 2 kh'y + (k^ - m^a^)b^ = 0,
whence
1, , . b^k
Eliminating k between these results, we find the equation of the
locus of the midpoints of the parallel chords of slope m :
X, § 226] ELLIPSE AND HYPERBOLA 217
(9) . y = -^x.
If m = tan a is the slope of any diameter of the ellipse (1),
the slope of the conjugate diameter is
mj = tan cti = -•
ma^
The diameter conjugate to this diameter of slope m^ has there-
fore the slope
_ 6^ _ ^' _
\ mo?)
i.e. it is the original diameter of slope m (Fig. 92). In other
words, either one of the diameters of slopes m and m^ is conjugate
to the other ; each bisects the chords parallel to the other.
226. Tangents Parallel to Diameters. Among the parallel
lines of slope m, y = mx 4- Jc, there are two tangents to the
ellipse, viz. (§ 221) those for which
7c = ± VmM + ^,
their points of contact lie on (and hence determine) the conju-
gate diameter. This is obvious geometrically; it is readily
verified analytically by. showing that the coordinates of the
intersections of the diameter of slope — li^/ma^ with the
ellipse (1) satisfy the equations of the tangents of slope m, viz.
y = mx ± ^m^a^ -f 6^.
The tangents at the ends of the diameter of slope m must of
course be parallel to the diameter of slope m-^. The four tan-
gents at the extremities of any two conjugate diameters thus
form a circumscribed parallelogram (Fig. 92).
The diameter conjugate to either axis of the ellipse is the
other axis ; the parallelogram in this case becomes a rectangle.
218
PLANE ANALYTIC GEOMETRY [X, § 227
227. Diameters of a Hyperbola. For the hyperbola the
same formulas can be derived except that ¥ is replaced
throughout by — 11^. But the geometrical interpretation is
somewhat different because a line y = mx meets the hyperbola
(3) in real points only when m < b/a.
Fig. 93
The solution of the simultaneous equations
y = 7nx,
gives :
b'^x'^
ay = a^b^
x = ±
ab
V62
y=±
mob
m^a^
Vb'
7n^a^
These values are real if m<b/a and imaginary if m>b/a
(Fig. 93). In the former case it is evidently proper to call the
distance PQ between the real points of intersection a diameter
of the hyperbola ; its length is
PQ = 2 VS^+7^ = 2 «* ^i^+MT.
If m>b/a, this quantity is imaginary; but it is customary to
speak even in this case of a diameter, its length being defined
as the real quantity
^ rn^a^ — b^
By this convention the analogy between the properties of the
ellipse and hyperbola is preserved.
X, § 228] ELLIPSE AND HYPERBOLA 219
228. Conjugate Diameters of a Hyperbola. Two diameters
of the hyperbola are called conjugate if their slopes 7n, mi are
such that
mrrii = —
One of these lines evidently meets the curve in real points, the
other does not.
If m < b/a, the line y = mx, as well as any parallel line,
meets the hyperbola (3) in two real points, and the locus of the
midpoints of the chords parallel to y = mx is found to be the
diameter conjugate to y — mx, viz.
y = miX = — - X.
ma^
If m > b/a, the coordinates a^, yi and ajg, 2/2 of the intersec-
tions of y=zmx with the hyperbola are imaginary; but the
arithmetic means ^ (X1 + X2), ^(?/i + ?/2) ^i'^ real, and the locus
of the points having these coordinates is the real line
b'
y = miX = — X.
ma^
It may finally be noted that what was in § 227 defined as
the length of a diameter that does not meet the hyperbola
in real points is the length of the real diameter of the hyper-
bola
• -^ + ^' = 1;
d? b""
two such hyperbolas are called conjugate.
220
PLANE ANALYTIC GEOMETRY [X, § 229
229. Parameter Equations. Eccentric Angle. Just as the
parameter equations of the circle x"^ -\- y"^ = o? are (§ 194) :
ic = a cos ^, y = a sin 0,
so those of the ellipse (1) are
ic = a cos dy y=h sin d,
and those of the hyperbola (3) are
a: = a sec ^, y =h tan 6.
In each case the elimination of the parameter $ (by squaring
and then adding or subtracting) leads to the cartesian equation.
The angle 6, in the case of the
circle, is simply the polar angle of
the point P (x, y). In the case of the
ellipse, as appears from Fig. 94
(compare § 212), 6 is the polar angle
not of the point P {x, y) of the ellipse,
but of that point Pi of the circum-r
scribed circle which has the same
abscissa as P, and also of that point
Pg of the inscribed circle which has the same ordinate as P.
This angle 6 = xOP^ is called the eccentric angle of the point
P (a;, y) of the ellipse.
In the case of the hyperbola the eccentric angle 6 determines
the point P(x, y) as follows (Fig. 95). Let a line through
inclined at the angle 6 to the trans-
verse axis meet the circle of radius
a about the center at A, and let the
transverse axis meet the circle of
radius h about the center at B. Let
the tangent at A meet the transverse
axis at A' and the tangent at B meet
the line OA at B'. Then the parallels to the axes through^'
and B' meet at P.
Fig. 94
Fig. 05
X, § 230] ELLIPSE AND HYPERBOLA * 221
230. Area of Ellipse. Since any ellipse of semi-axes a, b
can be regarded as the projection of a circle of radius a,
inclined to the plane of the ellipse at an angle € such that
cos € = b/a, the area A of the ellipse is ^ = vd^ cos c = -n-ab.
EXERCISES
1. Find the tangents to the ellipse x^ + iy^ = 16, which pass through
the following points :
(a) (2, V3), (b) (-3,iV7), (c) (4,0), (d) (-8,0).
\ 2. Find the tangents to the hyperbola 2 x^ — S y^ = IS, which pass
through the following points :
(a) (-6, 3V2), (&) (-3,0), (c) (4, -V5), (d) (0,0).
,-^ 3. Find the intersections of the line x — 2y = 7 and the hyperbola
x^-y^ = 5.
4. Find the intersections of the line Sx + y — 1 = and the ellipsQ
x^ + 4y^ = 65.
. ; ^'5. For what value of k will the line y = 2x + khe 3, tangent to the
hyperbola ic2-4y2-4 = 0?
-^ 6. For what values of m will the line y=7nx + 2 be tangent to the
ellipse x2 + 4 2/2 _ 1 = ?
7. Find the conditions that the following lines are tangent to the hy-
perbola x2/a2 - 1/2/62 = 1 .
(a) Ax + By -{- C = 0, (b) xcos^ + y sin p =p.
8. Are the following points on, outside, or inside the ellipse ^2+4 y2=4p
(«) (1,1), (b) (I, -i), (c) (-i, -I).
9. Are the following points on, outside, or inside the hyperbola
9x2-2/2 = 9? (^a) (f, -I), (6) (1.35,2.15), (c) (1.3,2.6).
~^ 10. Find the difference of the eccentric angles of points at the extremi-
ties of conjugate diameters of an ellipse.
11. Show that conjugate diameters of an equilateral hyperbola are
equal.
f- 12. Show that an asymptote is its own conjugate diameter.
- 13. Show that the segments of any line between a hyperbola and its
asymptotes are equal.
- 14. Find the tangents to an ellipse referred to its axes which have
equal, intercepts.
222 PLANE ANALYTIC GEOMETRY [X, § 230
15. What is the greatest possible number of normals that can be drawn
from a given point to an ellipse or hyperbola ?
16. Show that tangents drawn at the extremities of any chord of an
ellipse (or hyperbola) intersect on the diameter conjugate to the chord.
17. Show that lines joining the extremities of tlie axes of an ellipse
are parallel to conjugate diameters.
18. Show that chords drawn from any point of an ellipse to the ex-
tremities of a diameter are parallel to conjugate diameters.
19. Find the product of the perpendiculars let fall to any tangent from
the foci of an ellipse (or hyperbola).
20. The earth's orbit is an ellipse of eccentricity .01677 with the sun
at a focus. The mean distance (major semi-axis) between the sun and
earth is 93 million miles. Find the distance from the sun to the center
of the orbit.
21. Find the sum of the squares of any two conjugate semi-diameters
of an elUpse. Find the difference of the squares of conjugate semi-diam-
eters of a hyperbola.
22. Find the area of the parallelogram circumscribed about an ellipse
with sides parallel to any two conjugate diameters.
23. Find the angle between conjugate diameters of an ellipse in terms
of the semi-diameters and semi-axes.
24. Express the area of a triangle inscribed in an ellipse referred to
its axes in terms of the eccentric angles of the vertices.
25. The circle which is the locus of the intersection of two perpendicu-
lar tangents to an ellipse or hyperbola is called the director-circle of the
conic. Find its equation : {a) For the ellipse. (&) For the hyperbola.
26. Find the locus of a point such that the product of its distances
from the asymptotes of a hyperbola is constant. For what value of this
constant is the locus the hyperbola itself ?
27. Find the locus of the intersection of normals drawn at correspond-
ing points of an ellipse and the circumscribed circle.
28. Two points J., J5 of a line I whose distance is AB = a move along
two fixed perpendicular lines ; find the path of any point P of I.
CHAPTER XI
CONIC SECTIONS — EQUATION OF SECOND DEGREE
PART I. DEFINITION AND CLASSIFICATION
231. Conic Sections. The ellipse, hyperbola, and parabola
are together called conic sections, or simply conies, because
the curve in which a right circular cone is intersected by any
plane (not passing through the vertex) is an ellipse or hyper-
bola according as the plane cuts only one of the half-cones or
both, and is a parabola when the plane is parallel to a gener-
ator of the cone. This will be proved and more fully dis-
cussed in §§ 239-243.
232. General Definition. The three conies can also be
defined by a common property in the plane : the locus of a point
for ivhich the ratio of its distances from a fixed point and from
a fixed line is constant is a conic, viz. an ellipse if the constant
ratio is less than one, a hyperbola if
the ratio is greater than one, and a
parabola if the ratio is equal to one.
We shall find that this constant
ratio is equal to the eccentricity e — cja
as defined in § 215. Just as in the
case of the parabola for which the
above definition agrees with that of
§ 172, we shall call the fixed line d^ directrix, and the fixed
point jPj focus (Fig. 96).
223
y
L
^
__$
/f
i
D
X
Fi
<---,
K.-
--^
iL
y
d,
Fig. 96
224 PLANE ANALYTIC GEOMETRY [XI, § 233
233. Polar Equation. Taking the focus 2<\ as pole, the
perpendicular from Fi toward the directrix d^ as polar axis,
and putting the given distance F^D = q, we have FiP = r,
PQ = q — r cos <j>, r and <^ being the polar coordinates of any
point P of the conic. The condition
to be satisfied by the point P, viz.
FiP/PQ==e, i.e. F^P^e-PQ becomes,
therefore,
e(g — rcos <^),
whence r =
1 4- e cos <^ Fia. 96
y
L
'%
__«
/f
D
X
Fi
c— -,
k-
iL
?
d,
This then is the polar equation of a conic if the focus is taken
as pole and the perpendicular from the focus toward the directrix
as polar axis.
It is assumed that the distance q between the fixed point
and fixed line is not zero; the ratio e, i.e. the eccentricity of
the conic, may be any positive number.
234. Plotting the Conic. By means of this polar equation
the conic can be plotted by points when e and q are given.
Thus, for <^ = and <^ = tt, we find eq/{l -\- e) and eq/{l — e) as
the intercepts F^A^ and F1A2 on the polar axis ; A^, A2 are the
vertices. For any negative value of cf> (between and — tt)
the radius vector has the same length as for the same positive
value of <fi. The segment LL' cut off by the conic on the per-
pendicular to the polar axis drawn through the pole is called
the latus rectum; its length is 2 eg. Notice that in the ellipse
and hyperbola, i.e. when e ^1, the vertex Ai does not bisect
the distance FiD (as it does in the parabola), but that
F^Ai/A^D = e.
XI, § 236]
CONIC SECTIONS
225
If in Fig. 96, other things being equal, the sense of the
polar axis be reversed, we obtain
Fig. 97. We have again F^P= r ; but
the distance of P from the directrix
di is QP = q -\- r cos <f), so that the
polar equation of the conic is now :
._ ^1
1 — e cos <f>
y
P.
Q
~L^
J)
1 \
aA \
di
^ gr-U- >
^
Fig. 97
235. Classification of Conies. For e = 1, the equations of
§§ 233-234 reduce to the equations of the parabola given in
§§ 172, 173. It remains to show that for e < 1 and e > 1
these equations represent respectively an ellipse and a hyper-
bola as defined in §§ 204, 207.
To show this we need only introduce cartesian coordi-
nates and then transform to the center^ i.e. to the midpoint
between the intersections ^i, A^ of the curve with the polar
axis.
236. Transformation to Cartesian Coordinates. The equa-
tion of § 233,
T — e{q — r cos <^)
becomes in cartesian coordinates, with the pole F^ as origin
and the polar axis as axis Ox (Fig. 96) :
VaJ^ -\-y^= e{q — a;),
or rationalized :
(1 - e2).'c2 + 2 e V + / = e'^'.
The midpoint O between the vertices A-^, A^ at which the
curve meets the axis Ox has, by § 234, the abscissa
this also follows from the cartesian equation, with 2/ = 0.
226 PLANE ANALYTIC GEOMETRY [XI, § 237
237. Change of Origin to Center. To transform to paral-
lel axes through this point we have to replace x by
X — e^q/(l — e^) ; the equation in the new coordinates is there-
fore
and this reduces to
r'
i.e.
g2g2 ^2^2
(1 - 6^)2 l-e2
If e < 1 this is an ellipse with semi-axes
1 - e^ vnr^'
if e > 1 it is a hyperbola with semi-axes
238. Focus and Directrix. The distance c (in absolute value)
from the center O to the focus F^ is, as shown above, for the
ellipse „
c = — ^- = ae,
1 — e^
for the hyperbola
e — \
The distance (in absolute value) of the directrix from the
center is for the ellipse, since g = a(l — e^)/e = a/e — ae :
and for the hyperbola, since q = ae — a/e :
OD = c-q = ae-ae-\-- = --
e e
XI, §238] CONIC SECTIONS 227
It is clear from the symmetry of the ellipse and hyperbola
that each of these curves has two foci, one on each side of the
center at the distance ae from the center, and two directrices
whose equations are a; = ± a/e.
EXERCISES
1. Sketch the following conies :
2 + 3 COS 2 + cos 1 — 2 cos
2. Sketch the following conies and find their foci and directrices :
(a) ic2 + 4 1/2 = 4, (ft) 4 x2 + 1/2 _ 4^
(c) a:2 _ 4 ^2 :^ 4^ (ri) 4x2 - 2/2 = 4,
(e) 16 x2 + 25 2/2 = 400, (/) 9 a;2 - 16 2/2 = 144,
(^) 9 a;2 - 16 y^ + 144 = 0, (/t) x2 - 1/2 = 2.
3. Show that the following equations represent ellipses or hyperbolas
and find their centers, foci, and directrices :
(a) x2 + 32/2-2x+6?/ + l =0, (6) 12x2 - 41/2 - 12x - 9 = 0,
(c) 5x2 + y2 + 20x + 15 = 0, {d) 5x2-42/2 + 8?/ + 16 = 0.
4. Find the length of the latus rectum of an ellipse and a hyperbola
in terms of the semi-axes.
5. Show that the intersections of the tangents at the vertices with
the asymptotes of a hyperbola lie on the circle about the center passing
through the foci.
6. Show that when tangents to an ellipse or hyperbola are drawn
from any point of a directrix the line joining the points of contact passes
through a focus.
7. From the definition (§ 232) of an ellipse and hyperbola, show that
the sum and difference respectively of the focal radii of any point of the
conic is constant.
8. Find the locus of the midpoints of chords drawn from one end of :
(a) the major axis of an ellipse ; (&) the minor axis.
9. The eccentricity of an ellipse with one focus and corresponding
directrix fixed is allowed to vary; show that the locus of the ends of the
minor axis is a parabola.
10. Find the locus of § 232 when the fixed point lies on the fixed line.
228
PLANE ANALYTIC GEOMETRY [XI, § 239
239. The Conies as Sections of a Cone. As indicated by
their name the conic sections, i.e. the parabola, ellipse, and
hyperbola, can be defined as the curves in which a right circu-
lar cone is cut by a plane (§ 231).
In Figs. 98, 99, 100, Fis the vertex of the cone, ^ CVC' = 2 a
the angle at its vertex ; OQ indicates the cutting plane, CVC
that plane through the axis of the
cone which is perpendicular to the
cutting plane. The intersection
OQ of these two planes is evidently
an axis of symmetry for the conic.
The conic is a parabola, ellipse,
or hyperbola, according as OQ is
parallel to the generator VC of the
cone (Fig. 98), meets VC at a point
O' belonging to the same half-cone
as does O (Fig. 99), or meets FO'
at a point 0' of the other half-cone (Fig. 100).
COQ be called p, the conic is
Fig. 98
If the angle
a parabola if /3 = 2 a (Fig. 98),
an ellipse if ^ > 2 a (Fig. 99),
a hyperbola if ^ < 2 a (Fig. 100).
In each of the three figures CO represents the diameter 2 r
of any cross-section of the cone {i.e. of any section at right
angles to its axis). We take O as origin, OQ as axis Ox, so
that (Fig. 98) OQ. = x, QP=y are the coordinates of any point
P of the conic.
As QP is the ordinate of the circular cross-section CPC'P'
we have in each of the three cases :
y2^Qp2^CQ'QC\
XI, § 241]
CONIC SECTIONS
229
240. Parabola. In the first case (Fig. 98), when y8 = 2 a so
that OQ is parallel to VC, the expression
X OQ OQ ^
is constant, i.e. the same at whatever distance from the vertex
we may take the cross-section CPC'P'. For, QO is equal to
the diameter OB = ^r^ of the cross-section through 0, and
CQ/OQ = CC'I VC = 2 r/r esc « = 2 sin a.
Hence, denoting the constant r^ sin a by p we have
CQ
OQ
QC = 4 ?o sin a =4p.
The equation of the conic in this case, referred to its axis OQ
and vertex 0, is therefore
y^ = 4:px.
Notice that as p = Tq sin a the focus is
found as the foot of the perpendicular
from the midpoint of OB on OQ.
241. Ellipse. In the second case
(Fig. 99), i.e. when ^ > 2 a, if we put
Oa = 2a,
it can be shown that
f ^ QP'
x{2a-x) OQ-QO'
Fig. 99
is constant. For we have QP^ = CQ • QC and from the tri-
angles CQO, QCa, observing that ^ QaC = fi-2a:
9Sl = si")g QC' ^ sin(^-2ct)
OQ sin(i7r-a)' QO' sin(^7r + a)'
230
whence
PLANE ANALYTIC GEOMETRY [XI, § 241
QP' _ sin )8sin(/3-2a)
OQ • qa
cos^ a
an expression independent of the position of the cross-section
CO.
Denoting this positive constant by h^, we find the equation
y^ = k'^x(2 a — x),
(x-ay ^ y' ^^
i.e.
ikaf
which represents an ellipse, with semi-axes a, ka and center
(a, 0).
242. Hyperbola. In the third case
(Fig. 100), proceeding as in the second
and merely observing that now
qO' = -{2a + x\
we find the equation
y'^ = k^x{2a-\-x),
I.e.
(x-ha)'
(fca)
which represents a hyperbola, with
semi-axes a, ka and center (—a, 0).
Fig. 100
243. Limiting Cases. As the conic is an ellipse, hyperbola,
or parabola according as /8 > 2 a, < 2 a, or = 2 a, it appears
that ih.Q parabola can be regarded as the limiting case of either
an ellipse or a hyperbola whose center (the midpoint of OCy)
is removed to infinity.
On the other hand, if in the second case, /? > 2 a (Fig. 99),
XI. §2431 CONIC SECTIONS 231
we let p approach tt, or if in the third case, p <2 a (Fig. 100),
we let p approach 0, the cutting plane becomes in the limit a
tangent plane to the cone. • It then has in common with the
cone the points of the generator VC, and .these only. A single
straight line can thus appear as a limiting case of an ellipse or
hyperbola.
Finally we obtain another class of limiting cases, or cases of
degeneration, of the conies if, in any one of the three cases,
we let the cutting plane pass through the vertex V of the
cone. In the first case, (3 = 2 a, the cutting plane is then tan-
gent to the cone so that the parabola also may degenerate into
a single straight line. In the second case, ^ > 2 a, if /8 ^ tt,
the ellipse degenerates into a single point, the vertex V of the
cone. In the third case, /3 < 2 a, if /? ^ 0, the hyperbola de-
generates into two intersecting lines.
The term conic section, or coiiic, is often used as including
these limiting cases.
EXERCISES
1. For what value of /S in the preceding discussion does the conic be-
come a circle ? .
2. A right circular cylinder can be regarded as the limiting case of a
right circular cone whose vertex is removed to infinity along its axis
while a certain cross-section remains fixed. The section of such a cylin-
der by a plane is in general an ellipse ; in what case does it degenerate
into two parallel lines ?
3. The conic sections were originally defined (by the older Greek
mathematicians, in the time of Plato, about 400 b.c.) as sections of a
cone by a plane at right angles to a generator of the cone ; show that the
section is a parabola, ellipse, or hyperbola according as the angle 2 a at
the vertex of the cone is = | tt, < | tt, > | tt.
4. Show that the spheres inscribed in a right circular cone so as to
touch the cutting plane (Figs. 98, 99, 100) touch this plane at the foci of
the conic.
232 PLANE ANALYTIC GEOMETRY [XI, § 244
PART II. REDUCTION OF GENERAL EQUATION
244. Equations of Conies. We have seen in the two pre-
ceding chapters that hy selecting the coordinate system in a con-
venient way the equation of a parabola can be obtained in the
simple form .
y^=z4:px,
that of an ellipse in the form
a-'^b^-^'
and that of a hyperbola in the form
a^ ¥
When the coordinate system is taken arbitrarily, the carte-
sian equations of these curves will in general not have this
simple form ; but they will always be of the second degree.
To show this let us take the common definition of these curves
(§ 232) as the locus of a point whose distances from a fixed
point and a fixed line are in a constant ratio. With respect to
any rectangular axes, let x^ , 2/1 be the coordinates of the fixed
point, ax -{-by -\- c = the equation of the fixed line, and e the
given ratio. Then by §§9 and 56 the equation of the locus is
or, rationalized :
{x - x,y + {y - y,y = -^ (ax -hby-h c)\
a^ -j- 0^
It is readily seen that this equation is always of the second
degree; i.e. that the coefiicients of a;^, y"^, and xy cannot all
three vanish.
XI, § 246] EQUATION OF SECOND DEGREE 233
245. Equation of Second Degree. Conversely, every eq\ia-
tion of the second degree, i.e. every equation of the form (§ 79)
(1) Ax" + 2 Hxy -^ By"" + 2 Gx-\-2 Fy ■\- C = 0,
where A, H, B are not all three zero, in general represents a
conic. More precisely, the equation (1) may represent an
ellipse, a hyperbola, or a parabola; it may represent two
straight lines, different or coincident ; it may be satisfied by
the coordinates of only a single point; and it may not be
satisfied by any real point.
Thus each of the equations
a^ - 3 / = 0, xy =
evidently represent two real different lines ; the equation
ic2_2a; + l =
represents a single line, or as it is customary to say, two coin-
cident lines ; the equation
a;2 + ?/' =
represents a single point, while
is satisfied by no real point and is sometimes said to represent
an "imaginary ellipse."
The term conic is often used in a broader sense (compare § 243)
so as to include all these cases ; it is then equivalent to the
expression "locus of an equation of the second degree.'^
It will be shown in the present chapter how to determine
the locus of any equation of the form (1) with real coefficients.
The method consists in selecting the axes of coordinates so as
to reduce the given equation to its most simple form.
246. Translation of Axes. The transformation of the
equation (1) to its most simple form is very easy in the par-
ticular case ichen (1) contains no term in xy, i.e. when H = 0.
Indeed it suffices in this case to complete the squares in x and y
and transform to parallel axes.
234 PLANE ANALYTIC GEOMETRY [XI, § 246
Two cases may be distinguished:
(a) 11=0, A =^ Oj B =^0, so that the equation has the form
(2) Ax"" -\- By^ -}- 2 Gx + 2 Fy + C= 0.
Completing the squares in x and y (§ 80), we obtain an equation
of the form
A {X - hf -\-B{y- kf = K,
where ^ is a constant ; upon taking parallel axes through the
point {h, k) it is seen thatxthe locus is an ellipse, or a hyper-
bola, or two straight lines, or a point, or no real locus, accord-
ing to the values of A, B, K.
(h) H=0, and either ^= or J.=0, so that the equation is
(3) Ax'' + 2Gx + 2Fy-{- 0=0, or By' -h2Gx -\-2Fy + G=0.
Completing the square in x or y, we obtain
(x-hy=p(y-k), or (y -kf = q{x-h)',
with (h, k) as new origin we have a parabola referred to vertex
and axis, or two parallel lines, real and different, coincident, or
imaginary.
It follows from this discussion that the absence of the term in
xy indicates that, in the case of the ellipse or hyperbola, its axes,
in the case of the parabola, its axis and tangent at the vertex, are
parallel to the axes of coordinates.
EXERCISES
1. Reduce the following equations to standard forms and sketch the
loci : *
(a) 2 2/2 _ 3 a; + 8 1/ + 11 = 0, (b) x^ + ^y^ - 6x + iy + 6 = 0,
(c) 6 x2 + 3 ?/2 - 4 a: + 2 y + 1 = 0, (d) x^ - 9y^ - 6x + ISy = 0,
(e) 9 a;2 + 9 2/2 - 36 x+6 ?/+ 10 = 0, (/) 2 j:^ - iy"^ + 4 x + 4y - 1 = 0,
(9) x2 + i/2_2x + 22/-H3 = 0, (h) 3x2 - 6x + y + 6 = 0,
(0 x2 - ?/2 _ 4 X - 2 ?/ + 3 = 0, ( j) 2 x2 - 5 X + 12 = 0,
(A;) 2 x2 - 5 X -}- 2 = 0, (0 y-^ - 4 y + 4 = 0.
XI, §247] EQUATION OF SECOND DEGREE
235
2. Find the equation of each of the following conies, determine the
axis perpendicular to the given directrix, the vertices on this axis (by
division-ratio), the lengths of the semi-axes, and make a rough sketch
in each case :
(a) with x — 2 = as directrix, focus at (6, 3), eccentricity | ;
(i!>) with 3x-|-4y— 6 = 0as directrix, focus at (5, 4) , eccentricity | ;
(c) with X — ?/ — 2 = 0as directrix, focus at (4, 0), eccentricity |.
3. Find the axis, vertex, latus rectum, and sketch thfe parabola with
focus at (2, — 2) and 2a: — 3 y — 5 = as directrix (see Ex. 2).
4. Prove the statement at the end of § 244.
5. Find the equation of the ellipse of major axis 5 with foci at (0, 0)
and (3, 1).
247. Rotation of Axes. If the right angle xOy formed by
the axes Ox, Oy be turned about the origin through an
angle d so as to take the new position x^Oy^ (Fig. 101), the
relation between the old coordinates OQ = x, QP = y of any
point P and the new coordinates OQi^x^, QiP=yi of the
same point P are seen from the figure to be
< x = Xi cos — yi sin 0,
[ y = x^ sin + .Vi cos 6.
By solving for x^ , y^ , or again from Fig. 101, we find
j x'l = X cos 6 -\-y sin 9,
\y^ = — X ^\n 6 + y cos 6.
If the cartesian equation of any curve referred to the axes
(4)
(4')
236
PLANE ANALYTIC GEOMETRY [XI, § 247
Ox, Oy is given, the equation of the same curve referred to the
new axes Ox^ , Oyi is found by substituting the values (4) for
X, y in the given equation.
248. Translation and Rotation. To transform from any
rectangular axes Ox, Oy (Fig. 102) to any other rectangular
y
1
y,.
k
h
\ \
1 X
jt
Fia. 102
axes OxX^ , O^y-^ , we have to combine the translation 00^
(§ 13) with the rotation through an angle 6 (§ 247).
This can be done by first transforming from Ox, Oy to the
parallel axes Oyx\ O^y' by means of the translation (§ 13)
x = x^ -\-h,
y = y'-\- ^,
and then turning the right angle x'Oiy' through the angle
= x'OiXi , which is done by the transformation (§ 247)
x' = Xi cos 6 — yi sin 6,
2/' = iCi sin + yi cos 6.
Eliminating x', y', we find
x = XiC0s6 — 2/i sin 6 -{-h,
y = Xi sin 0-\-yi cos $ + lc.
The same result would have been obtained by performing
first the rotation and then the translation.
It has been assumed that the right angles xOy and x^Oy^ are
superposable ; if this were not the case, it would be necessary
to invert ultimately one of the axes.
(5)
XI, § 248] EQUATION OF SECOND DEGREE 237
EXERCISES
1. Find the coordinates of each of the following points after the axes
have been rotated about the origin through the indicated angle :
(a) (3, 4), ^T. (&) (0, 5),i7r.
(c) (-3, 2), <? = tan-i|. (d) (4,-3),^^-
2. K the origin is moved to the point (2, -r- 1) and the axes then
rotated through 30"^, what will be the new coordinates of the following
points?
(a) (0,0). (6) (2,3). (c) (6,-1).
3. Find the new equation of the parabola y^ = i ax after the axes have
been rotated through : (a) ^tt , (b) ^tt , (c) tt .
— 4. Show that the equation x^ + y'^ = a^ is not changed by any rotation
of the axes about the origin. Why is this true ?
5. Find the center of the circle {x— a)^ + y^ —a'^ after the axes have
been turned about the origin through the angle Q. What is the new
equation ?
- 6. For each of the following loci rotate the axes about the origin
through the indicated angle and find the new equation :
/(a) x2-i/2 + 2=0, Itt. (6) x^-y'^ = a\\ir.
I (c) 2/ = mx + 6, = tan-i m. (d) 12x^ - 7 xy - 12y^ = 0, d = t&n-^l-
Oi
7. Through what angle must the axes be turned about the origin so
that the circle x^-^-y^ — Sx + iy — 6 = will not contain a linear term
in x?
8. Suppose the right angle XiOyi (Fig. 101) turns about the origin at
a uniform rate making one complete revolution in two seconds. The
coordinates of a point with respect to the moving axes being (2, 1), what
are its coordinates with respect to the fixed axes xOy at the end of :
(a) i sec. ? (b) f sec. ? (c) 1 sec. ? (d) 1^ sec. ?
9. In Fig. 101, draw the line OP, and denote Z QOP by <f>. Divide
both sides of each of the equations (4) by OP and show that they are
then equivalent to the trigonometric formulas for cos (^ + 0) and
sin (d + <p).
238 PLANE ANALYTIC GEOMETRY [XI, § 249
249. Removal of the Term in xy. The general equation
of the second degree (1), § 245, when the axes are turned about
the origin through an angle ^ (§ 247), becomes :
A (a^i cos 6 — yi sin fff
+ 2 H(x^ cos d~ 2/i sin 6) {x^ sin O + yi cos 6)
+ J5(a;i sin (9 + 2/1 cos ^)2
+ 2 G{Xy^ cos d — yi sin $)
+ 2 F{x^ sin ^ + 2/i cos ^) + (7= 0.
This is an equation of the second degree in x^^ and y^ in
which the coefficient of x^y^ is readily seen to be
— 2^cos^sind + ^JBsin^cos^ + 2^(cos2(9-sin2^)
= {B- A) sin 2 ^ + 2 fi^cos 2 6.
It follows that if the axes be turned about the origin
through an angle 6 such that
(JS -^) sin 2 ^H- 2 ITcos 2 ^ = 0,
i.e. such that r
2H
(6) tan 2^
A-B'
the equation referred to the new axes will contain no term in
x^y^ and can therefore be treated by the method of § 246.
According to the remark at the end of § 246 this means
that the new axes Oa^i, Oyi, obtained by turning the original
axes Ox, Oy through the angle found from (6), are parallel
to the axes of the conic (or, in the case of the parabola, to the
axis and the tangent at the vertex).
The equation (6) can therefore be used to determine the
directions of the axes of the conic; but the process just indicated
is generally inconvenient for reducing a numerical equation of
the second degree to its most simple form since the values of
cos and sin 6 required by (4) to obtain the new equation are
in general irrational.
XI, §250] EQUATION OF SECOND DEGREE 239
EXERCISES
1. Through what angle must the axes be turned about the origin to
remove the term in xy from each of the following equations ?
(a) 3;:c2+2\/3a;?/+?/2_3a;+4?/-10=0. (6) x;^ + 2y/Ixy + 1 y'^-\^ = 0.
(c) 2x2- 3a;?/ + 2^/2 + x- 2/ +7=0. {d)xy = 2a'^.
2. Reduce each of the following equations to one of the forms in § 244 :
(a) xy = -% (6) 6 x2 - 5 xy - 6 2/2 = 0.
(c) 3x2-10x^ + 32/2 + 8 = 0. {d) 13x2 - lOxy + 13^2 _ 72 = o.
250. Transformation to Parallel Axes. To transform the
general equation of the second degree (1), § 245, to parallel
axes through any point (x^, y^), we have to substitute (§ 13)
x = x' + Xq, y=y' + yo,
the resulting equation is
Ax'' + 2 Hxy 4- By'' + 2 (Ax, + Hy, + G) a/
+ 2(J7a^o + 52/o + i^)/ + C" = 0,
where the new constant term is
(7) e = Ax,' + 2Hx,y,^By,' + 2Gx,-^2Fy,-\-a
It thus appears that after any trarislation of the coordinate
system :
(a) the coefficients of the terms of the second degree remain
unchanged ;
(b) the new coefficients of the terms of the first degree are
linear functions of the coordinates of the new origin ;
(c) the new constant term is the result of substituting the
coordinates of the new origin in the left-hand member of the
original equation.
240 PLANE ANALYTIC GEOMETRY [XI, § 251
251. Transformation to the Center. The transformed equa-
tion will contain no terms of the first degree, i.e. it will be of
the form _, -
(8) Ax"' + 2 H^y' + By"' -^ C = 0,
if we can>gelect the new origin {x^^ y^) so that
.gs / Ax,-{-Hy,+ G = 0,
^^ Hx, + By, + F^O.
This is certainly possible whenever
A H
and we then find :
no^ X - FH-GB ^ OH-FA
^ ^ "^ AB-H^' -^^ AB- 11^
As the equation (8) remains unchanged when x', y' are
replaced by — x\ — y', respectively, the new origin so found is
the center of the curve (§ 224). The locus is therefore in
this case a central conic, i.e. an ellipse or a hyperbola; but it
may reduce to two straight lines or to a point (see § 254). It
might be entirely imaginary, viz. if ^= ; but the case when
11=0 has already been discussed in § 246.
We shall discuss in § 256 the case in which AB — H^ = 0.
252. The Constant Term and the Discriminant. The cal-
culation of the constant term C can be somewhat simplified
by observing that its expression (7) can be written
C =(Ax,-{- H7j, + G)x,-h(Hx, + By, + F)y,-\- Gx,-\- Fy,-^ C,
i.e., owing to (9),
(11) C'=:Gx,-\-Fy,-^a
If we here substitute for x^, y^ their values (10) we find :
GFII - G^B + FGH - F^A -f- ABC - H'C
C
AB-H'
XI, § 253] EQUATION OF SECOND DEGREE 241
The numerator, which is called the discriminant of the equa-
tion of the second degree and is denoted by D, can be written
in the form of a symmetric determinant, viz.
A H G
D= H B F '
G F C
If we denote the cofactors of this determinant by the corre-
sponding small letters, we have
^0 — J ?/o — ' ^ —
C C C
Notice that the coefficients of the equations (9), which deter-
mine the center, are given by the first two rows of Z>, while the
third row gives the coefficients of C" in (11).
253. Homogeneous Function of Second Degree. The nota-
tion for the coefficients in the equation of the second degree arises from
the fact that the left-hand member of this equation can be regarded as
the value for 2; = 1 of the general homogeneous function of the second
degree, viz.
/(a;, y, z) = Ax'^ + By'^ -\- Cz^ + 2 Fyz + 2Gzx-{-2 Hxy.
If in this function x alone be regarded as variable while y and z are
treated as constants, the derivative with respect to x is
fj =2{Ax-\-Hy + Gz)',
if y alone, or z alone, be regarded as variable, we find similarly
fy' = 2{Hx + By + Fz),
f^' = 2{Gx + Fy+Cz).
These partial derivatives of the homogeneous function /(x, y, z) with
respect to a;, ?/, 2, respectively, are linear homogeneous functions of aj, y, z^
and it is at once verified that
i.e. the homogeneous function of the second degree is equal to half the sum
of the products of its partial derivatives by x, y, z.
A H
G
H B
F
G F
C
242 PLANE ANALYTIC GEOMETRY [XI, § 253
The left-hand members of the equations (0) are IfJixQ, yo , 1),
i/i/'C^o, Vqi !)• Hence the equations for the center can he obtained by
differentiating /(x, y, 0), or what amounts to the same, the left-hand
member of the equation of the second degree, with respect to x alone and
y alone.
The symmetric determinant
D
formed of the coefificients of ^/x', \fy -, \fz is called the discriminant of
f(x, y, z) ; and this is also the discriminant of the equation of the second
degree (§252). As f= i(fjx + fy'y -\-f,'z) andfJ{xo, yo, 1) = 0,
fy'(oco , yo, 1) = it follows that
C =f(xo , 1/0 , 1) = lf^'(xo , yo , 1) = Gxo -{-Fyo + C.
^ 254. Straight Lines. After transforming to the center, i.e.
obtaining the equation (8), we must distinguish two cases
according as G' = or C'=^0. The condition C' = means
by (7) that the center lies on the locus ; and indeed the homo-
geneous equation
represents two straight lines through the new origin (a^o , 2/0)
(§ 59). The separate equations of these lines, referred to
the new axes, are found by factoring the left-hand member.
As we here assume (§ 251) that AB — H^=^0, and H^O, the
lines can only be either real and different, or imaginary. In
the latter case the point (a;„ , y^) is the only real point whose
coordinates satisfy the original equation.
255. Ellipse and Hyperbola. If C =^0 we can divide (8)
by — C so that the equation reduces to the form
(12) ax' + 2hxy-\-by- = l.
This equation represents an ellipse or a hyperbola (since we
assume h=^0). The axes of the ellipse or hyperbola can be
found in magnitude and direction as follows.
XI, §255] EQUATION OF SECOND DEGREE
243
Fig. 103
If an ellipse or hyperbola, with its center, be given graphi-
cally, the axes can be constructed by inter-
secting the curve with a concentric circle
and drawing the lines from the center to
the intersections; the bisectors of the
angles between these lines are evidently
the axes of the curve (Fig. 103).
The intersections of the curve (12) with
a concentric circle of radius r are given by
the simultaneous equations
aa;2 H- 2 lixy -f h]f- = 1, ^-{-if^r^'^
dividing the second equation by r^ and subtracting it from the
first, we have
(13) ^a-iy + 2/10^7/ -f ^6 -iy^ = 0.
This homogeneous equation represents two straight lines
through the origin, and as the equation is satisfied by the
coordinates of the points that satisfy both the preceding equa-
tions, these lines must be the lines from the origin to the inter-
sections of the circle with the curve (12). If we now select r
(14)
(14-)
a —
so as to mak e thejbw o lines (13) coincid e, they will evidently
coincide with one or the other of the axes of the curve (12).
The condition for equal roots of the quadratic (13) in y/x is
This equation, which is quadratic in l/r^ and can be written
determines the lengths of the axes. If the two values found for
ir are both positive, the curve is an ellipse ; if one is positive
'-(aH-6)i + a6-7i2 = 0,
244 PLANE ANALYTIC GEOMETRY [XI, § 255
and the other negative, it is a hyperbola ; if both are negative,
there is no real locus.
Each of the two values of 1/r^ found from (14'), if substi-
tuted in (13), makes the left-hand member, owing to (14), a
complete square. Tlie equations of the axes are therefore
\a-h^±yjf>-^y = 0,
or, multiplying by Va — l/r^ and observing (14) :
a ]x -h hy = 0.
256. Parabola. It remains to discuss the case (§ 251) of the
general equation of the second degree.
Ax'' + 2 Hxy + By^ + 2Gx + 2 Fy +0 = 0,
in which we have ^^ _ jj2 _ q
This condition means that the terms of the second degree form
a perfect square :
Ax"" + 2 Hxy + By^ = (VAx + VSyy.
Putting V^ = a and V^ = 6 we can write the equation of the
second degree in this case in the form
(1 5) (ax + byf = -2Gx-2Fy-a
If G and F are both zero, this equation represents two parallel
straight lines, real and different, real and coincident, or im-
aginary according as (7 < 0, C = 0, (7 > 0.
If G and F are not both zero, the equation (15) can be inter-
preted as meaning that the square of the distance of the point
(x, y) from the line
(16) ax-}-by =
is proportional to the distance of (cc, y) from the line
(17) 2Gx + 2Fy-^C=0.
Hence if these lines (16), (17) happen to be at right angles, the
XI, § 256] EQUATION OF SECOND DEGREE 245
locus of (15) is Si parabola, having the line (16) as axis and the
line (17) as tangent at the vertex.
But even when the lines (16) and (17) are not at right angles
the equation (15) can be shown to represent a parabola. For
if we add a constant k within the parenthesis and compensate
the right-hand member by adding the terms 2 aJcx -f- 2 bky + 7c^,
the locus of (15) is not changed ; and in the resulting equation
(18) (ax + by + kf = 2(ak - G)x -f 2(bk - F)y -{-k^-C
we can determine k so as to make the two lines
(19) ax + by-^k = 0,
(20) 2(ak - G)x + 2{bk -F)y + k^-C=0
perpendicular. The condition for perpendicularity is
a{ak - G) -\-b{bk - F) = 0,
whence
(21) k^^^±^.
With this value of k, then, the lines (19), (20) are at right
angles ; and if (19) is taken as new axis Ox and (20) as new
axis Oy^ the equation (18) reduces to the simple form
y^ = px.
The constant p, i.e. the latus rectum of the parabola, is found
by writing (18) in the form
f ax 4- &y + ^ \_
2 V(afc - Gf + i^k - Fy- 2{ak - G)x + 2{bk~F)y-\-k''- (7 .
«'+&' * 2V(ak-Gy + (bk-Fy
hence
Substituting for k its value (21) we can reduce it to
^ 2(aF-bG)
{a'-\-¥)^
246 PLANE ANALYTIC GEOMETRY [XI, § 256
EXERCISES
1. Find the equation of each of the following loci after transforming
to parallel axes through the center :
(a) Sx^-4xy-y'^-Sx-iy + 7 = 0.
(6) 6 x^ + 6 xy -\- y^ + 6 X - 4: y — 6 = 0.
(c) 2 x^ -\- xy - 6 y^ — 7 X — 7 y -\- 5 = 0.
(d) x'^ - 2 xy - y^ -\- i X - 2y - 8 = 0.
2. Find that diameter of the conic Sx^ — 2xy—4:y'^+6x—4:y -^-2=0
(a) which passes through the origin, (&) which is parallel to each co-
ordinate axis.
3. For what values of k do the following equations represent straight
lines ? Find their intersections.
ia) 2x^ - xy-Sy'^-6x + 19y + k = 0.
(6) kx^ + 2 xy -{- y^ - X - y - 6 = 0.
(c) S x:^ - 4 xy + ky^ + S y - S = 0.
(d) x^-^2y^ + 6x-4y + k = 0.
4. Show that the equations of conjugate hyperbolas x^/a^—y^/b'^= ±1
and their asymptotes x^/a^—y'^/b^ = 0, even after a translation and rota-
tion of the axes, will differ only in the constant terms and that the con-
stant term of the asymptotes is the arithmetic mean between the constant
terms of the conjugate hyperbolas.
5. Find the asymptotes and the hyperbola conjugate to
2x^ — xy - 15y-^ + X+ 19y + 16=0.
6. Find the hyperbola through the point (—2, 1) which has the lines
2x — y+l = 0, 3x4-2?/ — 6 = as asymptotes. Find the conjugate
hyperbola.
7. Show that the hyperbola xy = a^ is referred to its asymptotes as
coordinate axes. Find the semi-axes and sketch the curve. Find and
sketch the conjugate hyperbola.
8. The volume of a gas under constant temperature varies inversely
as the pressure (Boyle's law), i.e. vp = c. Sketch the curve whose ordi-
nates represent the pressure as a function of the volume for different
values of c ; e.g. take c = 1, 2, 3.
9. Sketch the hyperbola (x — a)(y — b) = c^ and its asymptotes. In-
terpret the constants a, b, c geometrically.
XI, §256] EQUATION OF SECOND DEGREE 247
10. Sketch the hyperbola xy-\-Sy — 6 = and its asymptotes.
11. Find the center and semi-axes of the following conies, write their
equations in the most simple form, and sketch the curves :
^(a) 6 x^ - 6 xy + 5 y^ + 12y/2 X - W2y + 8 = 0.
^ (6) x2 - 6\/8 xi/ - 6 ^/2 - 16 = 0. (c) x'^ -{- xy -{- y^ - Sy + Q = 0.
(d) 13x^-QV3xy + 7y^-M = 0.
^ (e) 2 x2 - 4 X2/ + ?/2 4- 2 X - 4 ?/ - f = 0.
^ (/) 3 x2 + 2x2/ + 2/2 + 6x + 4 ?/ + I = 0.
12. Sketch the following parabolas :
(a) x2 _ 2V3xy + 3 «/2 - 6 V3 x -6y = 0.
(&) x2 - 6 xy + 9 «/2 - 3 X + 4y - 1 = 0.
13. Show that the following combinations of the coefficients of the
general equation of the second degree are invariants (i.e. remain un-
changed) under any transformation from rectangular to rectangular axes :
(a) A + B. (6) AB - H^. (c) (A - ^)2 + 4 iI2.
14. Show that x2 + y^ = a^ represents a parabola. Sketch the locus.
15. Find the parabola with x + y = as directrix and (^ a, | a) as
focus.
16. Let five points A, B, C, D, E be taken at equal intervals on a
line. Show that the locus of a point P such that AP ■ EP = BP • DP is
an equilateral hyperbola. (Take G as origin.)
17. The variable triangle AQB is isosceles with a fixed base AB.
Show that the locus of the intersection of the line AQ with the perpen-
dicular to QB through B is an equilateral hyperbola.
18. Let ^ be a fixed point and let Q describe a fixed line. Find the
locus of the intersection of a line through Q perpendicular to the fixed
line and a line through A perpendicular to AQ.
19. Find the locus of the intersection of lines drawn from the extrem-
ities of a fixed diameter of a circle to the ends of the perpendicular
chords.
20. Show by (14'), §255, that if the equation of the second degree
represents an ellipse, parabola, hyperbola, we have, respectively,
^S - If 2 ^ 0, = 0, < 0. .
CHAPTER XII
HIGHER PLANE CURVES
PART I. ALGEBRAIC CURVES
257. Cubics. It has been shown (§ 30) that every equation
of the first degree,
H- a^x + 6i2/ = 0,
represents a straight line; and (§ 245) that every equation of
the second degree,
Oo
+ a^x + biy
+ a^"^ 4- h^y + C22/2 = 0,
either represents a conic or is not satisfied by any real points.
The locus represented by an equation of the third degree,
4- a^x^ + h^y -f- c^"^
+ a^a? -f h^'^y + c^xy"^ + d^^= 0,
I.e. the aggregate of all real points whose coordinates x, y satisfy
this equation, is called a cubic curve.
Similarly, the locus of all points that satisfy any equation of
the fourth degree is called a quartic curve; and the terms quintic,
sextic, etc., are applied to curves whose equations are of the
Jifthj sixth, etc., degrees.
Even the cubics present a large variety of shapes; still
more so is this true of higher curves. We shall not discuss
such curves in detail, but we shall study some of their properties.
248
XII, §258] ALGEBRAIC CURVES 249
258. Algebraic Curves. The general form of an algebraic
equation of the .nth degree in x and y is
+ a^x -{-b^y
(1) + a^^ -f- b^xy + Cojy^
-\-a^-{- b^x^y 4- c^y"^ + d^
+ a^x"" + b^x^-^y + ...4- Kxtj^-'^+ l^y"" = 0.
The coefficients are supposed to be any real numbers, those in
the last line being not all zero. The number of terms is not
more than 1 + 2 + 3 + ... +(n + 1) = i(n + l)(n + 2).
If the cartesian equation of a curve can be reduced to this
form by rationalizing and clearing of fractions, the curve is
called an algebraic curve of degree n.
An algebraic curve of degree n can be intersected by a
straight line,
Ax-\- By-h C=0,
in not more than n points. For, the substitution in (1) of the
value of y (or of x) derived from the linear equation gives an
equation in x (or in y) of a degree not greater than n ; this
equation can therefore have not more than n roots, and these
roots are the abscissas (or ordinates) of the points of intersec-
tion.
We have already studied the curves that represent the poly-
nomial function
y=ao+ aiX-{-a^-\ hotna^";
such a curve is an algebraic curve, but it is readily seen by
comparison with the preceding equation that this equation is
of a very special type, since it contains no term of higher de-
gree than one in y. Such a curve is often called a parabolic
curve of the nth degree.
250
PLANE ANALYTIC GEOMETRY [XII, § 259
259. Transformation to Polar Coordinates. The cartesian
equation (1) is readily transformed to polar coordinates by sub-
stituting
X = r cos <^, y = r sin <^ ;
it then assumes the form :
+ (aj cos 4* -\-hi sin <^)?'
(2) -f (as cos^ <\>-{-h.2 cos <^ sin <J!) + Cg sin^ <l>)r^
+ (ttg cos^ <^ H- &3 cos^ <^ sin <^ -h Cg cos <j> sin^ <^ + c^s sin' <^)r^
+ (a„ cos" <^ + &„ cos"~^ <^ sin <j> +
+fc„cos<^sin"-^ <^-f/^ sin'* </>)?•»
= 0.
If any particular value be assigned to the polar angle <^, this
becomes an equation in r of a
degree not greater than n. Its
roots ri, r^,'" represent the in-
tercepts OPi, OP2, " (Fig. 104)
made by the curve (2) on the line
y = tan <^ • x. Some of these
roots may of course be imaginary,
and there may be equal roots. Fig. 104
260. Curve through the Origin. The equation in r has at
least one of its roots equal to zero if, and only if, the constant
term ao is zero. Thus, the necessary and sufficient condition that
the origin he a point of the curve is aQ = 0.
This is of course also apparent from the equation (1) which
is satisfied by ic = 0, 2/ = if, and only if, ao = 0.
261. Tangent Line at Origin. The equation (2) has at
least two of its roots equal to zero if ao = and ai cos <^ +
61 sin <f> = 0. If ai and bi are not both zero, the latter condition
XII, § 263]
ALGEBRAIC CURVES
251
can be satisfied by selecting the angle <^ properly, viz. so that
tan<^ = -^.
The line through the origin inclined at this angle <^ to the
polar axis is the tangent to the curve at the origin (Fig. 105).
Its cartesian equation is 2/ = tan <^'X — — (a^/h^x, i.e.
(3) a^x H- h^y = 0.
Thus, if tto = while Oi , by are not both zero, the curve has
at the origin a single tangent ; the origin is therefore called
a simple, or ordinary, point of the curve.
In other words, if the lowest terms in
the equation (1) of an algebraic curve
are of the first degree, the origin is a
simple point of the curve, and the equa-
tion of the tangent at the origin is ob-
tained by equating to zero the terms of
the first degree. Fig. 105
262. Double Point. The condition aicos <^ + ^i sin <^ =
necessary for two zero roots is also satisfied if «! = and &i = ;
indeed, it is then satisfied whatever the value of <^. Hence, if
a^ = 0, % = 0, 61 = 0, the equation (2) has at least two zero
roots for any value of <^. If in this case the terms of the
second degree in (1) do not all vanish, the curve is said to
have a double point at the origin. Thus, the origin is a double
l)oint if, and only if, the loivest terms in the equation (1) are of
the second degree.
263. Tangents at a Double Point. The equation (2) will
have at least three of its roots equal to zero if we have ao = 0,
ttj = 0, 61 = and
Oa cos'^ <^ 4- 62 cos <^ sin <^ + Cg sin^ <^ = 0.
252
PLANE ANALYTIC GEOMETRY [XII, § 263
K a^, 63, C2 are not all zero, we can find two angles satisfying
this equation which may be real and different, or real and
equal, or imaginary. The lines drawn at thjese angles (if real)
through the origin are the tangents at the double point.
Multiplying the last equation by 7^ and reintroducing carte-
sian coordinates we obtain for these tangents the equation
(4)
tta^J^ + b^y -\- c^y^ = 0.
Thus, if the loivest terms in the equation (1) are of the second
degree^ the origin is a double point, and these terms of the second
degree equated to zero represent the tangents at the origin.
264. Types of Double Point, (a) If the two lines (4) are
real and different, the double point is
called a node or crunode ; the curve then
has two branches passing through the
origin, each with a different tangent
(Fig. 106). ^
(b) If the lines (4) are coincident, i.e.
if ttg^ + b<p:y + c^y"^ is a complete square, Fig. 106
the double point is called a cusp, or spinode; the curve then
has ordinarily two real branches tangent to
one and the same line at the origin (Fig. 107
represents the most simple case).
(c) If the lines (4) are imaginary, the
double point is called an isolated point, or
an acnode; in this case, while the coordi-
nates 0, of the origin satisfy the equation
of the curve, there exists about the origin
a region containing no other point of the
curve, so that no tangents can be drawn
through the origin (Fig. 108).
J^
FiQ. 107
Fig. 108
XII, §265] ALGEBRAIC CURVES 253
It should be observed that, for curves of a degree above
the third, the origiu in case (b) may be an isolated point ; this
will be revealed by investigating the higher terms (viz. those
above the second degree).
265. Multiple Points. It is readily seen how the reasoning
of the last articles can be continued although the investigation
of higher multiple points would require further discussion.
The result is this : If in the equation of an algebraic curve, when
rationalized and cleared of fractions, the lowest terms are of
degree k, the origin is a k-tuple point of the curve, and the tan-
gents at this point are given by the terms of degree k, equated
to zero.
To investigate whether any given point (xi , y^ of an alge-
braic curve is simple or multiple it is only necessary to trans-
fer the origin to the point, by replacing xhy x + x^^ and y by
y + Vij and then to apply this rule.
EXERCISES
1. Determine the nature of the origin and sketch the curves :
{a) y = x'^~-2x. (b) x^ = 4y-y\ (^c) {x + a)(y + a) = a"^.
(d) ?/2 = a;2(4-x). {e)y^ = 3^. (f) x^ + y'^ = xK
(g) y^ = x^ + 7?. (h) x^ - 3 axy -\-y^ = 0. (i) x*- y* + 6 xy^ = 0.
2. Determine the nature of the origin and sketch the curve (y—x^y=x^,
for: (a) n = l. (6) n = 2. (c) w = 3. (d) w = 4.
3. Locate the multiple points, determine their nature, and sketch the
curves :
(a) y^ = x{x + S)^. (b) (y-3)2 = a;-2. (c) (y ^ 1)^ = (x - S)\
(c?)y8=(x + l)(x-l)2.
4. Sketch the curve y'^={x — a)(x — b){x—c) and discuss the multi-
ple points when :
(a) 0<a<6<c. (6) 0<a<& = c. (c) 0<a = 6<c. {d) 0<a = b = c.
254 PLANE ANALYTIC GEOMETRY [XII, § 266
PART 11. SPECIAL CURVES
DEFINED GEOMETRICALLY OR KINEMATICALLY
266. Conchoid. A fixed point and a fixed line I, at the
distance a from O, being given, the radius vector OQ, drawn from
to every point Q of I, is produced by a segment QP= b of con-
stant length; the locus of P is called the conchoid of Nicomedes.
For as pole and the perpendicular to I as polar axis the
equation of lis ri = a/ cos <^ ; hence that of the conchoid is -X
If the segment QP be laid off in the opposite sense we obtain
the curve
r = b
cos <^
which is also called a conchoid. Indeed, these two curves
are often regarded as merely two branches of the same
curve. Transforming to cartesian coordinates and rationaliz-
ing, we find the equation
(«-a)2(a;2-t-2/2) = 6V,
which represents both branches. Sketch the curve, say for
b = 2 a, and for b = a/2, and determine the nature of the origin.
267. Limacon. If the line I be replaced by a circle and the
fixed point be taken 07i the circle, the locus of P is called
Pascal's limacon.
For as pole and the diameter of the circle as polar axis
the equation of the circle, of radius a, is r^ = 2 a cos <^ ; hence
that of the limaqon is : ^ ^^--^^
r = 2 a cos <f> -\- b. t/y^
XII, § 268]
SPECIAL CURVES
255
If h = 2a the curve is called the cardioid ; the equation
then becomes
r = 4 a cos^ ^ 4*.
Sketch the limaqons for 6 = 3 a, 2 a, a ; transform to car-
tesian coordinates and determine the character of the origin.
268. Cissoid. 00' = a being a diameter of a circle, let any
radius vector drawn from meet the circle and its tangent at 0'
at the points Q, D, respectively; if on this radius vector we lay
off OR = QD, the locus of E is called the cissoid of Diodes.
With as pole and 00' as polar axis, we have
OD = a/cos <f), OQ = a cos <^ ;
the equation is therefore
= fi( cos A 1= a
\cos <j> J
_^ sin'<^ ^
cos (^'
or in cartesian coordinates
2 ^
Fig. 109
If instead of taking the difference of the radii vectores of the
circle and its tangent, we take their sum we obtain the so-called
companion of the cissoid,
r = a(cos <^ 4- sec <^),
I.e.
Sketch this curve.
2/2 = x'
2a — X
x — a
IL 269. Versiera. With the data of § 268, let us draw through
Q a parallel to the tangent, through R a parallel to the diameter ;
■ the locus of the point of intersection P of these parallels is
called the versiera (wrongly called the " witch of Agnesi ").
256
PLANE ANALYTIC GEOMETRY [XII, § 269
We have evidently with as origin and 00' as axis Ox
x = a cos^ <l>, y = o. tan <^,
whence eliminating <^ :
a?
x =
2/2 -+- a2
If we replace the tangent at 0' by any
perpendicular to 00' (Fig. 110), at the ^
distance h from 0, we obtain the curve
x = a cos^ <^, y = b tan <j>,
which reduces to the versiera for b = a.
Sketch the versiera, and the last curve for 6 = i a.
Fig. 110
270. Cassinian Ovals. Lemniscate. Two fixed points F„
F2 being given it is known that the locus of a point P is :
> VK"^
Fig. Ill
(a) a circle if FiP/F.,P = const. (Ex. 7, p. 90);
(6) an ellipse if F^P-^-F^P^z const. (§ 204) ;
(c) a hyperbola if i^iP- i?^2^= const. (§ 207).
The locus is called a Cassinian oval if JF\P • FgP = const. If
XII, § 271]
SPECIAL CURVES
257
we put FiF2 = 2a, the equation, referred to the midpoint
between F^ and F2 as origin and OF2 as axis Ox, is
lix + af 4- /] [.{X - af + 1/^ = T^'-
In the particular case when k = a^ the curve passes through
the origin and is called a lemniscate. The equation then re-
duces to the form
{x^ + y^y^2a\x^-y%
which becomes in polar coordinates
r" = 2a2 cos 2 <^.
Trace the lemniscate from the last equation.
271. Cycloid. The common cycloid is the path described by
any point P of a circle rolling over a straight line (Fig. 112).
If A be the point of contact of the rolling circle in any posi-
tion, the point of the given line that coincided with the point
P of the circle when P was point of contact, it is clear that
the length OA must equal the arc AP=a6, where a is the
radius of the circle, and 6= "^ACP the angle through which
the circle has turned since P was at O. The figure then shows
that, with O as origin and OA as axis Ox :
X = OQ = aO — a sin 0, y = a — a cos 6.
These are the parameter equations of the cycloid. The curve has
258
PLANE ANALYTIC GEOMETRY [XII, § 272
an infinite number of equal arches, each with an axis of sym-
metry (in Fig. 112, the line x = ird) and with a cusp at each
end. Write down the cartesian equation.
272. Trochoid. The path described by any point P rigidly
connected with the rolling circle is called a trochoid. If the
Fig. 113. —The Trochoids
distance of P from the center C of the circle is 6, the equations
of the trochoid are
x=^ad — b sin $, y = a — h cos 6.
Draw the trochoid for h = \a and f or 6 = | a.
273. Epicycloid. The path described by any point P of a
circle rolling on the outside of a fixed circle is called an epicy-
cloid (Fig. 114).
Let be the center, h the
radius, of the fixed circle, C the
center, a the radius, of the rolling
circle; and let Aq be that point
of the fixed circle at which the
describing point P is the point
of contact. Put A^OA = <^, ACP
= 6. As the arcs AAq and AP
are equal, we have
6<^ = ad.
Fig. 114
XII, § 274] ^ SPECIAL CURVES 259
With as origin and OAq as axis of x we have
a; = (a H- b) cos <f> + a sin [^ — (| rr — <^)],
2/ = (a 4- 6) sin <^ — « cos [^ — (|- TT — <^)],
i.e, x = (a-\-b) cos <^ — a cos <^,
2/ = (a + o) sm <^ — a sin — ! — <f>.
274. Hypocycloid. If the circle rolls on the inside of the
fixed circle, the path of any point of the rolling circle is called
a hypocycloid. The equations are obtained in the same way ;
they differ from those of the epicycloid merely in having a re-
placed by — a :
X = (b — a) cos <f> -\- a cos "~ ^ <^,
CL
y = (b — a) sin <^ — a sin ~ ^ <^.
Show that : (a) for b = 2a the hypocycloid reduces to a
straight line, and illustrate this graphically ; (6) for b = 4:a the
equations become
a;= 3 a cos <^-|- a cos 3 <f> — a cos' </>,
2/ = ^ <^ sin <^ — a sin 3 <^ = asin' <^,
whence x^-\-y^ = a^-^
sketch this four-cusped hypocycloid.
^ b EXERCISES
1. Sketch the following curves: (a) Spiral of Archimedes r = a<f>;.
(6) Hyperbolic spiral r(^ a ; (c) Lituus r^^ = a^.T^
.2. Sketch the following curves : (a) r = a sm<p ; (6) r = a cos ;
^cj) r = a sin 2 ; (^r = a cos 2 ; (e) r = a cos 3 ; (/) r = a sin 30 ;
(g) r = acos4 0; (T^ r = a sin 4 0.
3. Sketch with respect to the same axes the Cassinian ovals (§ 270)
for a = 1 and k = 2, 1.5, 1.1, 1, .75, .6, 0.
260 PLANE ANALYTIC GEOMETRY [XII, § 274
4. Let two perpendicular lines AB and CD intersect at 0. Through
a fixed point Q of AB draw any line intersecting CD at E. On this line
lay off in both directions from B segments BP of length OB. The locus
of P is called the strophoid. Find the equation, determine the nature of
O and Q, and sketch the curve.
5. Show that the lemniscate (§ 270) is the inverse curve of an equi-
lateral hyperbola with respect to a circle about its center.
6. Show that the strophoid (Ex. 4) is the curve inverse to an equilat-
eral hyperbola with respect to a circle about a vertex with radius equal
to the transverse axis.
7. Show that the cissoid (§ 268) is the curve inverse to a parabola
with respect to a circle about its vertex.
8. Find the curve inverse to the cardioid (§267) with respect to a
circle about the origin.
9. Transform the equation
a (s;2 + y2) ^ r^z
to polar coordinates, indicate a geometrical construction, and draw the
curve.
10. A tangent to a circle of radius 2 a about the origin intersects the
axes at T and 2^, find and sketch the locus of the midpoint P between T
and T'.
11. From any point Q of the line x = a draw a line parallel to the axis
Ox intersecting the axis Oy at C Find and sketch the locus of the foot
of the perpendicular from O on OQ.
12. The center of a circle of radius a moves along the axis Ox. Find
and sketch the locus of the intersections of this circle with lines joining
the origin to its highest point.
13. The center of a circle of radius a moves along the axis Ox. Find
and sketch the locus of its points of contact with the lines through the origin.
14. The center of a circle of radius a moves along the axis Ox. Its in-
tersection with the axis nearer the origin is taken as 'the center of another
circle which passes through the origin. Find and sketch the locus of the
intersections of these circles.
XII, § 276] TRANSCENDENTAL CURVES 261
PART III. SPECIAL TRANSCENDENTAL CURVES
275. The Sine Curve. The simple sine curve, y = sin x,
is best constructed by means of an auxiliary circle of radius
one. In Fig. 115, OQ is made equal to the length of the arc
OA = X ; the ordinate at Q is then equal to the ordinate BA of
the circle.
y
Fig. 115
Construct one whole j^enod of the sine curve, i.e. the portion
corresponding to the whole circumference of the auxiliary
circle ; the width 2 ^ of this portion is called the period of the
function sinx.
The simple cosine curve, y = cos x, is the same as the sine
curve except that the origin is taken at the point (^tt, 0).
The simple tangent curve, y = tan x, is derived like the sine
curve from a unit circle. Its period is tt.
276. The Inverse Trigonometric Curves. The equation
y = sin a; can also be written in the form
X = sin~^ y, or a; = arc sin y.
The curve represented by this equation is of course the same
as that represented by the equation y = sin x.
But if X and y be interchanged, the resulting equation
X = sin y, or y = sin~^ x, y — arc sin x,
represents the curve obtained from the simple sine curve by
reflection in the line y=x(^ 135).
262 PLANE ANALYTIC GEOMETRY [XII, § 276
Notice that the trigonometric functions sin x, cos x, tan x, etc.,
^re one-valued, i.e. to every value of x belongs only one value
of the function, while the inverse trigonometric functions sin~^ a;,
cos~^a7, tan~^a?, etc., are many-valued; indeed, to every value of
X, at least in a certain interval, belongs an infinite number of
values of the function.
EXERCISES
1. From a table of trigonometric functions, plot the curve y = sinx.
2. Plot the curve y = sinx by means of the geometric construction
of §275.
3 Plot the curve y = cosx (a) from a table ; (6) by a geometric con-
struction similar to that of § 275.
4. Plot the curve y = tan x from a table.
5. Plot each of the curves
(a) y = sm2 x. (d) y = sec x.
(6) y = 2 cos 3 a;. {e) y = ctn 2 x.
(c) y = 3 tan (x/2). (/) y = 2 tan 4 x.
6. Plot each of the curves
(a) y = sin-i x. (6) y = cos-i x. (c) y = tan-i x.
7. By adding the ordinates of the tw^o curves y = sin x and y = cos x,
construct the graph oi y = sin x + cos x.
8. Draw each of the curves
(a) y zzsiux + 2 cos x. (c) y = secx-{- tan x.
(6) y = 2 sin x + cos(x/2). (d) ?/ = sin x + 2 sin 2 x + 3 sin 3x.
9. The equation x = sin t, where t means the time and x means the
distance of a body from its central position, represents a Simple Harmonic
Motion. From the graph of this equation, describe the nature of the
motion.
277. Transcendental Curves. The trigonometric and in-
verse trigonometric curves, as well as, in general, the cycloids
and trochoids, are transcendental curves, so called because the
relation between the cartesian coordinates x, y cannot be ex-
pressed in finite form {i.e. without using infinite series) by
XII, § 279] TRANSCENDENTAL CURVES
263
means of the algebraic operations of addition, subtraction, mul-
tiplication, division, and raising to a power with a constant
exponent.
278. Logarithmic and Exponential Curves. Another very
important transcendental curve is the exponential curve
y = «^
and its inverse, the logarithmic
curve 1
y = log« X,
where a is any positive constant
(Fig. 116). A full discussion
of these curves can only be given
in the calculus. We must here
confine ourselves to some special
cases and to a brief review of the
fundamental laws of logarithms.
279. Definitions. The logarithm 6 of a number c, to the
base a (positive and ^ 1), is defined as the exponent b to which
the given base a must be raised to produce the number c
(§ 105) ; thus the two equations
a^ = c and b = log„ c
express exactly the same relation between b and c. It follows
that a'"""'' = c, whatever c.
If in the first law of exponents (§ 104), a^a'' = a^'^'^, we put
aP=Py a«= Q, a'+«=iV, so that PQ=N, we find since p=loga P,
q = loga Q,p + q = loga ]sr= log, pq -.
(1) log„PQ = log,P+log„Q.
Similarly we find from a^/a*^ = a^'" :
P
(2)
loga ^= log. P-l0g„Q.
264 PLANE ANALYTIC GEOMETRY [XII, § 279
If in the third law of exponents (§ 104), (a^y = a^", we put
.a** = P, aP" = M, so that P"" = M, we find since p = log„ P,
pn = log„ M:
(3) log„(P") = nlog,P.
These laws (1), (2), (3) of logarithms are merely the trans-
lation into the language of logarithms of the first and third
laws of exponents.
280. Napierian or Natural Logarithms. In the ordinary
tables of logarithms the base is 10, and for numerical calcula-
tions these common logarithms (Briggs' logarithms) are most
■convenient. In the calculus it is found that another system
of logarithms is better adapted to theoretical considerations ;
the base of this system is an irrational number denoted by e,
6 = 2.718281828 ...,
and the logarithms in this system are called natural logarithms
(or Napierian, or hyperbolic, logarithms).
281. Change of Base. Modulus. To pass from one system
of logarithms to another observe that if the same number N has
the logarithm p in the system to the base a and the logarithm
g in the system to the base b so that
a^ = N, p= log„ N, h" = N, q= log, N,
then q = logj, N= log^ a^=p log^, a,
by (3); i.e.
,(4) iogi^=log„^.logj,a.
Hence if the logarithms of the system with the base a are
known, those with the base b are found by multiplying the
logarithms to the base a by a constant number, logj,a.
Thus taking a = 10, b = e, we have
(4') log,iV^=log,o^^.log,10;
XII, §281] TRANSCENDENTAL CURVES 265
i.e. to find the natural logarithm of any number we have merely
to multiply its common logarithm by the number
log, 10 = 2.30258 509 ....
The reciprocal of this number,
M= — i— = 0.43429 448 • • .,
logg 10
i.e. the factor by which the natural logarithms must be multi-
plied to produce the common logarithms, is called the modulus
of the common system of logarithms.
In any system of logarithms, the logarithm of the base is
always equal to 1, by the definition of the logarithm (§ 279).
Hence, if in (4) we take iV"= &, we find
(5) log„6 .log,a = l.
In particular, with a = 10, 6 = e we have
(5') log.oe. log, 10 = 1;
i.e. the modulus M of the common logarithms is
Jf = —i— = logio e = 0.43429 448 ... .
log, 10
EXERCISES
1. From a table of logarithms of numbers, draw the curve y = logio x.
2. By multiplying the ordinates of the curve of Ex. 1 by 3, construct
the curve y = S logio x.
3. From the figure of Ex. 1, construct the curve y = 10* by reflection
of the curve of Ex. 1 in the line y = x.
4. Draw the curve y = ^ logio x by the process of Ex. 2. Show that it
represents the equation y = logioo x, since
y = logioo X = logioo 10 X logic x = ^ logic X.
5. Find logio 7 from a table. Construct the curve
y = logy X = logioa; ^ logio 7
by the process described in Ex. 2 and Ex. 4.
6. Given logic e = M= .43+, draw the curve
y = log, X = logic X ^ logic e.
266 PLANE ANALYTIC GEOMETRY [XII, § 282
PART IV. EMPIRICAL EQUATIONS
282. Empirical Formulas. In scientific studies, the rela-
tions between quantities are usually not known in advance,
but are to be found, if possible, from pairs of numerical values
of the quantities discovered by experiment.
Simple cases of this kind have already been given in §§ 15,
29. In particular, the values of a and h in formulas of the
type y = a-{-bx were found from two pairs of values of x and y.
Compare also § 34.
Likewise, if two quantities y and x are known to be connected
by a relation of the form y = a-\-bx-\- cx"^, the values of a, b, c
can be found from any three pairs of values of x and y. For,
if any pair of values of x and y are substituted for x and y
in this equation, we obtain a linear equation for a, b, and c.
Three such equations usually determine a, b, and c.
In general the coefficients a, b, c, •••, Z in an equation of the
^^ y z= a -\- bx -\- cx"^ -\- " ' -\- Ix""
can be found from any n + 1 pairs of values of x and y.
283. Approximate Nature of Results. Since the measure-
ments made in any experiment are liable to at least small
errors, it is not to be expected that the calculated values of
such coefficients as a, &, c, • • • of § 282 will be absolutely accu-
rate, nor that the points that represent the pairs of values of
X and y will all lie absolutely on the curve represented by the
final formula.
' To increase the accuracy, a large number of pairs of values
of X and y are usually measured experimentally, and various
pairs are used to determine such constants as a, &, c, --of § 282.
The average of all the computed values of any one such con-
stant is often taken as a fair approximation to its true value.
XII, § 284]
EMPIRICAL EQUATIONS
267
284. Illustrative Examples.
Example 1. A wire under tension is found by experiment to stretch
an amount I, in thousandths of an inch, under a tension T, in pounds, as
follows : —
T in pounds 10 15 20 25 30
I in thousandths of an inch . 8 12.5 15.5 20 23
Find a relation of the form I = kT (Hookers Law) which approx-
imately represents these results.
First plot the given data on squared paper, as in the adjoining figure.
dU
'~'
—
—
""
""■
"~*
■""
■""
25
/
/
<
./
/
/
20
^
/
/
/'
/
/
15
t
V(
)
,/
/
y
/
10
,/
/
/
J
/
^/
^
5
/
/
/
/
/
/
\
s
\
n
\
"i
7
2
5
.-"i
n
7s^
Fig. 117
Substituting ? = 8, T = 10 in ? = A:r, we find A; = .8. From I = 12.5,
T — 15, we find k = .833. Likewise, the other pairs of values of I and T
give, respectively, k = .775, k = .8, k= .767. The average of all these
values of A; is A: = .795 ; hence we may write, approximately,
I = .795 T.
268
PLANE ANALYTIC GEOMETRY [XII, § 284
This equation is represented by the line in Fig. 117 ; this line does not
pass through even one of the given points, but it is a fair compromise be-
tween all of them, in view of the fact that each of them is itself probably
slightly inaccurate.
Example 2. In an experiment with a Weston Differential Pulley
Block, the effort E^ in pounds, required to raise a load IF, in pounds, was
found to be as follows :
w
10
20
30
40
50
60
70
80
90
100
E
3i
4|
6i
n
9
101
12i
13f
15
161
Find a relation of the form E
with these data.
aW +h that approximately agrees
[Gibson]
These values may be plotted in the usual manner on squared paper.
They will be found to lie very -^
nearly on a straight line. If E
is plotted vertically, h is the in-
tercept on the vertical axis, and
a is the slope of the line ; both
can be measured directly in the
figure.
To determine a and h more
exactly, we may take various
points that lie nearly on the
line. Thus {E = Q\, pr=30)
and {E = 16^, W = 100) lie
nearly on a line that passes close
to all the points. Substituting in the equation E = aW -^ h ^e obtain
6| = 30a + 6, 16J = 100a+&
whence a = 0.146, h = 1.86. Hence we may take
E= 0.146 Tr+ 1.86
approximately. Other pairs of values of E and W may be used in like
manner to find values f or « and 6, and all the values of each quantity may
be averaged.
T
: ^'
^« -
1^
»^
.^
_ 1
i> r
10 - - -
V ~~ ~ "
^'
-^ ^
s* - -
5 - -jr--- - -
a^
y'
'. w
V 20 40
60 80 IDO.
Fig. 118
XII, §284] EMPIRICAL EQUATIONS 269
Example 3. If 6 denotes the melting point (Centigrade) of an alloy
of lead and zinc containing x per cent of lead, it is found that
X = % lead 40 50 60 70 80 90
^ = melting point .... 186° 205° 226° 250° 276° 304°
Find a relation of the form 6 = a -{• bx + cx^ that approximately expresses
these facts. [Saxelby]
Taking any three pairs of values, say (40, 186), (70, 250), (90, 304),
and substituting in d = a -\- bx -]- cx^ we find
186 = « + 40 6 + 1600 c,
260= a + 70 b + 4900 c,
304 = a + 90 6 + 8100 c,
whence a = 132, b = .92, c = .0011, approximately ; whence
e = 132 + .92x+ .0011x2.
Other sets of three pairs of values of x and y may be used in a similar
manner to determine «, 6, c ; and the resulting values averaged, as above.
EXERCISES
1. In experiments on an iron rod, the amount of elongation I (in thou-
sandths of an inch) and the stretching force p (in thousands of pounds)
were found to be {p = 10, l=S), (p = 20, Z = 15), (p = 40, Z = 31).
Find a formula of the. type l=k-p which approximately expresses these
data. Ans. k = .775.
2. The values 1 in. =2.5 cm. and 1 ft. =30.5 cm. are frequently
quoted, but they do not agree precisely. The number of centimeters, c,
in i inches is surely given by a formula of the type c = ki. Find k ap-
proximately from the preceding data.
3. The readings of a standard gas-meter S and those of a meter T being
tested on the same pipe-line were found to be (<S'=3000, r=0), (*9=3510,
T = 500), (S = 4022, T = 1000) . Find a formula of the type T= aS+ b
which approximately represents these data.
4. An alloy of tin and lead containing x per cent of lead melts at the
temperature d (Fahrenheit) given by the values (a: = 25%, ^ = 482°),
(x = 50%, d = 370°), (ic = 75%, d = 356°). Determine a formula of the
type 6 = a + bx + cx^ which approximately represents these values.
270 PLANE ANALYTIC GEOMETRY [XII, § 284
5. The temperatures d (Centigrade) at a depth d (feet) below the sur-
face of the earth in a mine were found to be <? = 100, 6 = 15.7° ; d = 200,
^=16.5 ; d=300, ^=17.4. Find a relation of the form d=a-\-bd between
e and d.
6. Determine a line that passes reasonably near each of the three
points !(2, 4), (6, 7), (10, 9). Determine a quadratic expression
y=a + hx-\-cx^ that represents a parabola through the same three points.
7. Determine a parabola whose equation is of the form y = a-}-bx-\-cx^
that passes through each of the points (0, 2.5), (1.5, 1.5), and (3.0, 2.8).
Are the values of «, &, c changed materially if the point (2.0, 1.7) is
substituted for the point (1.5, 1.5) ?
8. If the curve y = sinx is drawn with one unit space on the ic-axis
representing 60^, the points (0, 0), (^, J), (I2, 1) lie on the curve. Find a
parabola of the form y=a-\-bx-{-cx^ through these three points, and draw
the two curves on the same sheet of paper to compare them.
285. Substitutions. It is particularly easy to test whether
points that are given by an experiment really lie on a straight
line ; that is, whether the quantities measured satisfy an equa-
tion of the form y = a-\-bx. This is done by means of a trans-
parent ruler or a stretched rubber band.
For this reason, if it is suspected that two quantities x and
y satisfy an equation of the form
y = a + bx\
it is advantageous to substitute a new letter, say u, for x^ :
u = x^j y = a -{- bu
and then plot the values of y and u. If the new figure does
agree reasonably well with some straight line, it is easy to find
a and 6, as in § 284.
Likewise, if it is suspected that two quantities x and y are
connected by a relation of the form
2/ = a -f 6 • - or xy = ax-{-bf
X
it is advantageous to make the substitution u = 1/x.
XII, § 286] EMPIRICAL EQUATIONS 271
Other substitutions of the same general nature are often
useful.
In any case, the given values of x and y should he plotted first
unchanged, in order to see what substitution might he useful,
286. Illustrative Example. If a body slides down an inclined
plane, the distance s that it moves is connected with the time t after it
starts by an equation of the form s = kP: Find a value of k that agrees
reasonably with the following data :
s, in feet 2.6 10.1 23.0 40.8 63.T
t, in seconds 1 2 3 4 5
In this case, it is not necessary to plot the values of s and t themselves,
because the nature of the equation, s = kt'^^ is known from physics.
Hence we make the substitution t^ = u, and write down the supple-
mentary table :
s, in feet 2.6 10.1 23.0 40.8 63.7
w (or «2) 1 4 9 16 25
These values will be found to give points very nearly on a straight line
whose equation is of the form s = ku. To find k, we divide each value of
s by the corresponding value of u ; this gives several values of k :
k 2.6 2.525 2.556 2.55 2.548
The average of these values of k is approximately 2.556 ; hence we may
write s = 2.556 m, or s = 2.556 t^.
EXERCISES
1. Find a formula of the type u = kv^ that represents approximately
the following values : "
tt 3.9 15.1 34.5 61.2 95.5 137.7 187.4
t;12 34567
272 PLANE ANALYTIC GEOMETRY [XII, § 286
2. A body starts from rest and moves s feet in t seconds according to
the following measured values :
s, in feet 3.1 13.0 30.6 50.1 79.5 116.4
t, in seconds 5 1 15 2 2.5 3
Find approximately the relation between s and t.
3. The pressure p, measured in centimeters of mercury, and the volume
V, measured in cubic centimeters, of a gas kept at constant temperature,
were found to be :
145
155
165
178
191
L17.2
109.4
102.4
95.0
88.6
p
Substitute u for l/u, compute the values of m, and determine a relation
of the form p = ku; that is, p = k/v.
4. Determine a relation of the form y = a + bx^ that approximately
represents the values :
X 1 2 3 4 5 6 7
y 14.1 25.2 44.7 71.4 105.6 147.9 197.7
287. Logarithmic Plotting. In case the quantities y and x
are connected by a relation of the form
y = kx%
it is advantageous to take logarithms (to the base 10) on both
sides :
log y = log Tex"" = log k -{-n log x,
and then substitute new letters for log x and log y :
u = log Xj V — log y.
For, if we do so, the equation becomes
v = l -{- nu,
where I = log k.
XII, § 287]
EMPIRICAL EQUATIONS
273
If the values of x and y are given by an experiment, and if
u = log X and v = log y are computed, the values of u and v
should correspond to points that lie on a straight line, and the
values of I and n can be found as in § 284. The value of k
may be found from that of /, since log k= l.
Example 1. The amount of water A, in cu. ft. that will flow per
minute through 100 feet of pipe of diameter d, in inches, with an initial
pressure of 50 lb. per sq. in., is as follows :
d
1
1.5
2
3
4
6
A]
4.88
13.43
27.50
75.13
152.51
409.54
Fmd a relation between A and d.
Let u = \ogd, V = log A ; then the values of u and v are
u = \ogd . .
. 0.000
0.176
0.301
0.477
0.602
0.778
v = \ogA. .
. 0.688
1.128
1.439
1.876
2.183
2.612
:::::::::::: :::::::::::-5=:::::::
i .2 .3 4 .5
Fig. 119
j6 .7 .8
These values give points in the (u, v) plane that are very nearly on
a straight line ; hence we may write, approximately,
V = a+ bu,
where a and b can be determined directly by measurement in the figure,
T
274 PLANE ANALYTIC GEOMETRY [XII, § 287
or as in § 284. If we take the first and last pairs of values of u and v, we
find
.688 = a + 0,
2.612 = a + .778&.
Solving these equations, we find approximately, a = .688, b = 2.473,
and we may vvrrite
V = .688 + 2.473 u or log A = .688 + 2.473 log d.
Since .688 = log 4.88,
the last equation may be wrritten in the form
log A = log 4.88 + 2.473 log d
= log(4.88c?2-*78)
whence ^ = 4.88 (?2. 473.
Slightly different values of the constants may be found by using other
pairs of values of u and v.
288. Logarithmic Paper. Paper called logarithmic paper
may be bought that is ruled in lines whose distances, horizon-
tally and vertically, from one point (Fig. 120) are propor-
tional to the logarithms of the numbers 1, 2, 3, etc.
Such paper may be used advantageously instead of actually
looking up the logarithms in a table, as was done in § 287.
For if the given values be plotted on this new paper, the result-
ing figure is identically the same as that obtained by plotting
the logarithms of the given values on ordinary squared paper.
Example. A strong rubber band stretched under a pull of p kg.
shows an elongation of E cm. The following values were found in an ex-
periment :
p 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 6.0 6.0 7.0
E 0.1 0.3 0.6 0.9 1.3 1.7 2.2 2.7 3.3 3.9 5.3 6.9
[RiGGS]
If these values are plotted on logarithmic paper as in Fig. 120, it is evi-
dent that they lie reasonably near a straight line, such as that drawn.
XII, § 288]
EMPIRICAL EQUATIONS
275
By measurement in the figure, the slope of this line is found to be 1.6
approximately. Hence if m = log j? and v = log ^ we have
where I is a constant not yet determined ; whence
log^=:Z + 1.6 1ogp
or E = A.pi-6,
<n^
ltf_...
7:
.
t
-J- -
;'
^
/
/
/
" E = elongation in c
p = pull in kg.
/
3
m.
J
2
' Ez=.2
pi
.6
—
15
=
^
— _
!
'
/
2-:_
:::::i;i
: z_
f'
T J
)
fc _
/
_l
z
•z __
k = .3;z '^
7
.2----
=
7"~~
=
15
E
z2
Trrrrr
E
X
__
.^L
i iB .2 .5 .4 .5 .6 .7.8.91 15 2 "S 4 5 6 7 8 910'
Fig. 120. — Elongation of a Rubber Band
where Z = log A;. If j? = 1, JS' = A: ; from the figure, if p = 1, ^ = .3 ;
hence A: = .3, and
E = .3j!)l-6.
The use of logarithmic paper is however not at all essential ;
the same results may be obtained by the method of § 287.
276 PLANE ANALYTIC GEOMETRY [XII, § 288
EXERCISES
1. In testing a gas engine corresponding values of the pressure p^ meas-
ured in pounds per square foot, and the volume v, in cubic feet, were
obtained as follows : v = 7.14, p = 54.6 ; 7.73, 50.7 ; 8.59, 45.9. Find
the relation between p and v (use logarithmic plotting).
Ans. p = 387.6 v-^^, or pv^ = 387.6.
2. Expansion or contraction of a gas is said to be adiabatic when no
heat escapes or enters. Determine the adiabatic relation between pressure
p and volume v (Ex. 14) for air from the following observed values :
p = 20.54, V = 6.27 ; 25.79, 5.34 ; 54.25, 3.15.
Ans. pv^-'^ = 273.5.
3. The intercollegiate track records for foot-races are as follows,
where d means the distance run, and t means the record time :
d 100 yd. 220 yd. 440 yd. 880 yd. 1 mi. 2 mi.
t 0:09| 0:21^ 0:48 l:54f 4:15f 9:24|
Plot the logarithms of these values on squared paper (or plot the
given values themselves on logarithmic paper). Find a relation of the
form t = kd\ What should be the record time for a race of 1320 yd. ?
[See Kennelly, Popular Science Monthly, Nov. 1908.]
4. Solve the Example of § 288 by the method of § 287.
5. Each of the following sets of quantities was found by experiment.
Find in each case an equation connecting the two quantities, by §§ 287-
288.
(a) V
P
1
137.4
2
62.6
3
39.6
4
28.6
5
22.6
(6) u
V
12.9
63.0
17.1
27.0
23.1
13.8
28.5
8.5
3.0
6.9
(c) e
c
82^^
2.09
212°
2.69
390°
2.90
570°
2.98
750°
3.09
1100^
3.28
SOLID ANALYTIC GEOMETRY
CHAPTER XIII
COORDINATES
289. Location of a Point. The position of a point in three-
dimensional space can be assigned without ambiguity by giv-
ing its distances from three mutually rectangular planes, pro-
vided these distances are taken with proper signs according as
the point lies on one or the other side of each plane.
The three planes, each perpendicular to the other two, are
called the coordinate planes ; their common point (Fig. 121)
is called the origin. The three
mutually rectangular lines Ox,
Oy, Oz in which the planes in-
tersect are called the axes of
coordinates; on each of them
a positive sense is selected
arbitrarily, by affixing the
letter x, y, z, respectively.
The three coordinate planes,
Oyz, Ozx, Oxy, divide the whole
of space into eight compartments called octants. The first
octant in which all three coordinates are positive is also called
the coordinate trihedral.
If P', P", P'" are the projections of any point P on the
coordinate planes Oyz, Ozx, Oxy, respectively, then P'P=x,
P"P = y, P'"P= z are the rectangular cartesian coordinates of
277
/!
/
1
^y
z
?/"--
•
""-^
V
Q' y
Fig. 121
278
SOLID ANALYTIC GEOMETRY [XIII, § 289
P. If the planes through P parallel to Oyz, Ozx, Oxy intersect
the axes Ox, Oy, Oz in Q', Q", Q'", the point P is found from
its coordinates x, y, z by passing along the axis Ox through the
distance 0Q'= x, parallel to Oy through the distance Q'P"=yj
and parallel to Oz through the distance P"P=z, each of
these distances being taken with the proper sense.
Every point in space has three definite real numbers as coordi-
nates; conversely, to every set of three real numbers corresponds
one and 07ily one point.
Locate the points : (2, 3, 4), (- 3, 2, 0), (5, 0,-3), (0, 0, 4),
(0,-6,0), (-5, -8, -2).
290. Distance of a Point from the Origin. For the distance
OP—r (Fig. 121) of the point P{x, y, z) from the origin we
have, since OP is the diagonal of a rectangular parallelepiped
with edges OQ' =x, OQ" =y, OQ"' = z:
M
^
t
r
r = -\Qi9' + 2/^ + z^.
291. Distance between two Points. The distance between
the two points Pj (aJi,2/i>^i) ^^^ A
(^2 ) 2/2 J ^2) can be found if the coordi-
nates of the two points are given.
For (Fig. 123), the planes through P^
and those through P^ parallel to the
coordinate planes bound a rectangular
parallelepiped with P^Pi = d as di-
agonal ; and as its edges are
PiQ = x^-x,, PiR=y2-yi
we find
d = V(^2 - ^if + (2/2 - ViY + (^2 - ^if-
292. Oblique Axes. The position of a point P in space can also
be determined with respect to three axes not at right angles. The coor-
dinates of P are the segments cut off on the axes by planes through P
Fig. 122
PS = z.
XIII, § 292] COORDINATES 279
parallel to the coordinate planes. In what follows, the axes are always
assumed to be at right angles unless the contrary is definitely stated.
EXERCISES
1. What are the coordinates of the origin ? What can you say of the
coordinates of a point on the axis Ox ? on the axis Oij ? on the axis Oz ?
2. What can you say of the coordinates of a point that lies in the
plane Oxy ? in the plane Oyz ? in the plane Ozx ?
3. Where is a point situated when a; = ? when = 0? when
x = y = 0? when y = z'} when x = 2? when = — 3 ? when x = 1 ,
2/ = 2?
4. A rectangular parallelepiped lies in the first octant with three of
its faces in the coordinate planes, its edges are of length a, &, c, respec-
tively ; what are the coordinates of the vertices ?
5. Show that the points (4,3, 5), (2, -1,3), (0,1,7) are the
vertices of an equilateral triangle.
6. Show that the points (- 1, 1, 3), (— 2, — 1, 4), (0, 0, 5) lie on a
sphere whose center is (2, — 3, 1). What is the radius of this sphere ?
7. Show that the points (6, 2, - 5), (2, - 4, 7), (4, - 1, 1) lie on a
straight line.
8. Show that the triangle whose vertices are (a, 6, c) , (6, c, a) , (c, a, 6)
is equilateral.
9. What are the coordinates of the projections of the point (6, 3, — 8)
on the axes of coordinates ? What are the distances of this point from the
coordinate axes ?
10. What is the length of the segment of a line whose projections on
the coordinate axes are 5, 3, and 2 ?
11. What are the coordinates of the points which are symmetric to
the point (a, 6, c) with respect to the coordinate planes ? with respect to
the axes ? with respect to the origin ?
12. Show that the sum of the squares of the four diagonals of a rec-
tangular parallelepiped is equal to the sum of the squares of its edges.
280 SOLID ANALYTIC GEOMETRY [XIII, § 293
293. Projection. The projection of a point on a plane or
line is the foot of the perpendicular let fall from the point on
the plane or line. The projection of a rectilinear segment AB
on a plane or line is the intercept A'B' between the feet of the
perpendiculars AA', BB' let fall from A, B on the plane or
line. If a is one of the two angles made by the segment with
the plane or line we have
A'B' = AB cos a.
In analytic geometry we have generally to project a vector,
i.e. a segment with a definite sense, on an axis, i.e. on a line
with a definite sense (compare § 19). The angle a is then
understood to be the angle between the positive senses of
vector and axis (both being drawn from a common origin).
The above formula then gives the projection with its proper
sign.
Thus, the segment OP (Fig. 121) from the origin to any
point P(x, y, z) can be regarded as a vector OP. Its projec-
tions on the axes of coordinates are
the coordinates x, y, z of P. These
projections are also called the rec-
tangular components of the vector OP,
and OP is called the resultant of the
components OQ', OQ", OQ'", or also
of OQ', qP'", P"'P.
Similarly, in Fig. 123, if P^P^ be Fig. 123
regarded as a vector, the projections of this vector P^P^ on the
axes of coordinates are the coordinate differences x^ — x^,
2/2 — 2/i ) 2^2 — 2i . See § 298.
294. Resultant. The proposition of § 19 that tlie sum of
the projections of the sides of an open polygon on any axis is
<?"
r
XIII, § 295]
COORDINATES
281
equal to the projectioyi of the dosing side on the same axis and
that of § 20 that the projection of the resultant is equal to the
sum of the projections of its components are readily seen to hold
in three dimensions as well as in the plane. Analytically
these propositions follow by considering that whatever the
points P,{x^, 2/1, z^), P,(X2, y^, z^), ••• P„K , y^, z^) in space,
the sum of the projections of the vectors PyPi, P-iP^, ••• Pn-i^n
on the axis Ox is :
(x^-x,)-{-{x,-x,)-i- •.. -\-(x^-x^_i) = x^-Xij
where the right-hand member is the projection of the closing
side or resultant PiP„ on Ox. Any line can of course be taken
as axis Ox.
295. Division Ratio. Two points P-i{x^, y^ z{) and
Pi (^2 J 2/2 ) ^2) being given by their
coordinates, the coordinates x, y, z
of any point P of the line PyP^
can he found if the division ratio
P^P/P^p2 = k is known in ivhich
the point P divides the segment
P,P, (Fig 124).
Let Qi, Q, Qabe the projections
of Pj, P, P2 on the axis Ox-, as
Q divides Q1Q2 in the same ratio k in which P divides P1P2,
we have as in § 3 :
X =^ X-^ ~\~ K (3/2 — "^1/*
Similarly we find by projecting on Oy, Oz :
2/ = 2/1 + A: (2/2 - yO, z = Zi + k(z, - z^).
If k is positive, P lies on the same side of P^ as does Pg ; if
k is negative, P lies on the opposite side of Pj (§ 3).
Fig. 124
282 SOLID ANALYTIC GEOMETRY [XIII, § 296
296. Direction Cosines. Instead of using the cartesian
coordinates x, y, z to locate a point P (Fig. 125) we can also
use its radius vector r = OP, i.e. the length of the vector drawn
from the origin to the point, and its direction cosines, i.e. the
cosines of the angles a, /3, y, made
by the vector OP with the axes Ox,
Oy, Oz. We haye evidently
ic = f cos a, 2/ = r cos p, z = r cos 7.
As a line has two opposite senses
we can take as direction cosines jY V
of any line parallel to OP either Fig. 125
cos a, cos p, cos y, or — cos a, — cos ft, — cos y.
The direction cosines cos a, cos ft, cos y of a vector OP are
often denoted briefly by the letters Z, m, n, respectively, so
that the coordinates of P are
x—lr, y = mr, z = nr.
The direction cosines of any parallel line are then I, m, n
or —I, —m, — n.
297. Pythagorean Relation. The sum of the squares of the
direction cosines of any line is equal to one.
For, the equations of § 347 give upon squaring and adding
since 7? -\- y"^ -\- z^ = r"^ :
cos'^ ct + cos^ P + cos^ 7 = 1?
or
Z2 + m2 4- ^i' = 1 ;
and this still holds when I, m, n are replaced by —l,—m,— n.
Since this result is derived directly from the Pythagorean
Theorem of geometry, it may be called the Pythagorean Rela-
tion between the direction cosines. Notice that I, m, n can be
regarded as the coordinates of the extremity of a vector of
unit length drawn from the origin parallel to the line.
XIII, §297] COORDINATES 283
EXERCISES
1. Find the length of the radius vector and its direction cosines for
each of tlie following points : (5, - 3, 2); (- 3, - 2, 1); (- 4, 0, 8).
2. The direction cosines of a line are proportional to 1, 2, 3; find
their values.
3. A straight line makes an angle of 30*^ with the axis Ox and an
angle of 60° with the axis Oy ; what is the third direction angle ?
4. What is the direction of a line when Z = ? when Z = w = ?
5. What are the direction cosines of that line whose direction angles
are equal ?
6. What are the direction cosines of the line bisecting the angle
between two intersecting lines whose direction cosines are Z, w, n and I ,
to', w', respectively ?
7. Find the direction cosines of the line which bisects the angle
between the radii vectores of the points (3, — 4, 2) and (— 1, 2, 3).
8. Three vertices of a parallelogram are (4, 3, —2), (7, — 1, 4),
(—2, 1, — 4); find the coordinates of the fourth vertex (three solutions).
9. In what ratio is the line drawn from the point (2, — 5, 8) to the
point (4, 6,-2) divided by the plane Ozx ? by the plane Oxy ? At what
points does this line pierce these coordinate planes ?
10. In what ratio is the line drawn from the point (0, 5, 0) to the
point (8, 0, 0) divided by the line in the plane Oxy which bisects the
angle between the axes ?
11. Find the coordinates of the midpoint of the line joining the points
(4, — 3, 8) and (6, 5, — 9) . Find the points which trisect the same segment.
12. If we add to the segment joining the points (4, 1, 2) and (—2,
5, 7) a segment of twice its length in each direction, what are the coordi-
nates of the end points ?
13. Find the coordinates of the intersection of the medians of the tri-
angle whose vertices are Pi (xi , yi , Zx), Ti {xi^ yi, zt), Tz (xs , yz , zz).
14. Show that the lines joining the midpoints of the opposite edges
of a tetrahedron intersect and are bisected by their common point.
16. Show that the projection of the radius vector of the point
P(x, «/, z) on a line whose direction cosines are l\ to', w' is Vx -f- m^y + n^z.
284
SOLID ANALYTIC GEOMETRY [XIII, § 298
r
"298. Projections. Components of a Vector. If two points
-PiC^'ij Viy ^\) and P2('^"2) ?/2j ^2) are given by their coordinates,
the projections of the vector, P1P2 on
the axes, or what amounts to the
same, on parallels to the axes drawn
through Pj (Eig. 126), are evidently
(§ 293) : •
PyQ = X2-x,, P,R = ^2 - 2/1,
P^S = Z2 — Zi.
These projections, or also the vectors
PiQ, QN,NP2y are called the rectangular components of the
vector P1-P2 > or its components along the axes.
If d is the length of the segment PiPo , its direction cosines Z,
m, n are since P^Q is perpendicular to P^Q, P2R to P^E, P2S
to P,S:
Fig. 126
1 =
3/2
?/o — V,
These relations can also be written in the form :
X2—Xy ^ ^2 — .Vl ^ ^2 — '^! ^ fl
I m n
(li,m2,nt)
299. Angle between two Lines. Iftlie directions of two lines
are given by their direction cosines li , mi , n^ and I2 , ^2 , ng , the
angle ij/ between the two lines is given
by the formula
cos x|/ = I1I2 + mitn^2 -I- nin2-
For, drawing through the origin
two lines of direction cosines li , mi ,
ni and ^2 > ^2 > ^2 and taking on the jj/^ ^'
former a vector OPi of unit length, Fig- 127
the projection OP of OPi on the other line is equal to the
r
/
/ ^v^^ ^^l^miMii
£^>-'''^V y
5^
\Jy
XIII, § 3011 COORDINATES 285
cosine of the required angle ij/. On the other hand, OPi has
h) *^i) % ^s components along the axes ; hence, by § 294 :
cos {{/ = I1I2 + mimg + 711112.
Two intersecting lines (or any two parallels to them) make
two angles, say xf/ and ir — \p. But if the direction cosines of
each line are given, a definite sense has been assigned to each
line, and the angle between the lines is understood to be the
angle between these senses.
300. Conditions for Parallelism and for Perpendicularity.
If, in particular, the lines are parallel, we have either l^ == I2,
mi = m2 , Til = 712, or li = — 12, 771^ = — m2, tii = — Wj ; hence in
either case l,^ni,^7i.
This then is the condition of parallelism of two lines whose
direction cosines are ^1, m^, n^ and I2, m,, 712.
If the lines are perpendicular, i.e. if j/^=i7r, we have
cos 1/^ = 0; hence the condition of perpendicularity of two lines
whose direction cosines are li, mi, n^ and I2 , m2, 712 is
I1I2 + mim2 + ni7i2 = 0.
301. The formula of § 299 gives
sm2 xp = 1 - cos2 xf/ = 1 — (I1I2 4-mim2 + ^1^2)^.
As (§ 297) (Zi2 + mi2 + ni^)(l2^ + m2^ + n2'^)= 1, we can write this ex-
pression in the form
sin2 xl/ = ^^^ "^ *^^^ "^ ^^^ ^^^2 '^ ''^^'"^^ "*■ '*^^2
hh 4- mim2 + nin2 l-^ + m-^ + ^2^
which, by Ex. 3, p. 45, can also be expressed as follows :
h WI2I
The direction (I, m, w) perpendicular to two given different directions
(l\ , wi , ni) and (^2 , W2 , W2) is found by solving the equations (§ 300)
III + m\m + Jiiw = 0,
hi + wi2W + n2n = 0,
sin^i//
mi ?ii
•i
n\ h
+
+
m2 ^2
W2 ^2
286
whence
SOLID ANALYTIC GEOMETRY [XIII, § 301
{
m
n
mi
Wl
7ll h
h mi
W2
W2
W2 h
h W2
If we denote by k the common value of these ratios, we have
h mi
h Wi2
1 =
mi wi
W2 W2
Wl ii
W2 Z2
substituting these values in the relation (§ 297) P- + m^ + ^2 _ 1 and,
observing the preceding value of sin ^, we find :
l=±
mi ni
m^ 712
m =±
li mi
h Wl2
sin ^ sin \p
where \p is the angle between the given directions.
sin^
302. Three directions (Zi, wii, Wi), (Z2, wi2, ih), (Izi WI3, W3) are com-
planar, i.e. parallel to the same plane, if there exists a direction (Z, m, w)
perpendicular to all three. This will be the case if the equations
hi + wiim + Tiin = 0,
hi + W2m + n2n — 0,
?3? + wi3m +[?i3n =
have solutions not all zero ; hence the condition of complanarity
h mi ni
h mi 112.
h mz Wa
EXERCISES
1. Find the length and direction cosines of the vector drawn from the
point (5, —2, 1) to the point (4, 8, — 6) ; from the point (a, &, c) to the
point ( — a, —6, — c) ; from ( — a, —h, — c) to (a, ?), c).
2. Show that when two lines with direction cosines ?, w, 7i and
l\ w', 7i', respectively, are parallel, IV + wi>^' + wn' =±1. 2
3. Show that when two lines with direction cosines proportional to
a, 6, c, and a', 6', c', are perpendicular aa' + &&'+ cc' = ; and when the
lines are parallel a/ a' =h/h' = c/c'.
4. Show that the points (5, 2, -3), (6, 1, 4), (-2, -3, 6),
(—1, — 4, 13) are the vertices of a parallelogram.
XIII, § 303] COORDINATES 287
5. Show by direction cosines that the points (6, —3, 5), (8, 2, 2),
(4, —8, 8) lie in a line.
6. Find the angle between the vectors from (5, 8, — 2) to (—2, 6,-1)
and from (8, 3, 5) to (1, 1, -6).
7. Find the angles of the triangle whose vertices are (5, 2, 1),
(0,3, -1),(2, -1,7).
8. Find the direction cosines of a line which is perpendicular to two
lines whose direction cosines are proportional to 2, —3, 4, and 5, 2, —1,
respectively.
9. Derive the formula of § 299 by taking on each line a vector of unit
length, OPi and OP2, and expressing the distance P1P2 first by the
cosine law of trigonometry, then by § 292, and equating these expressions.
10. Find the rectangular components of a force of 12 lb. acting along
a line inclined at 60° to Ox and at 45° to Oy.
11. Find the resultant of the forces OPi, OP2, OP3, OP4 if the co-
ordinates of Pi, P2, P3, P4, with O as origin, are (3, —1, 2), (2, 2,-1),
(-1,2,1), (-2, 3, -4).
12. If any number of vectors, applied at the origin, are given by the
coordinates x, y, z of their extremities, the length of the resultant H is
\/(Sx)2 + {^yy^ + (Ss)"-^ (see Ex. 9, p. 21), and its direction cosines
are S xjB, S yjB, S zIB.
13. A particle at one vertex of a cube is acted upon by seven forces
represented by the vectors from the particle to the other seven vertices ;
find the magnitude (length) and direction of the resultant.
14. If four forces acting on a particle are parallel and proportional to
the sides of a quadrilateral, the forces are in equilibrium, i.e. their resultant
is zero. Similarly for any closed polygon.
303. Translation of Coordinate Trihedral. Let x, y, z be
the coordinates of any point P with respect to the trihedral
formed by the axes Ox, Oy, Oz (Fig. 128). If parallel axes
^i^ij ^lVu ^1% t)6 drawn through any point Oi(a, 6, c), and if
^j> 2/ij ^1 ^^^ the coordinates of P with respect to the new tri-
288
SOLID ANALYTIC GEOMETRY [XIII, § 303
hedral OxX{y^Zi, then the relations between the old coordinates
X, y, z, and the new coordinates Xy, y^, z^ of one and the same
point P are evidently
x = a-}-x^, y = b + 7ji, z = c + z^.
The coordinate trihedral has thus
been given a translation, represented
by the vector 00^^. This operation
is also called a transformation to
parallel axes through Oi- Fig. 128
. 304. Area of a Triangle. Any two vectors OPi, OP2 drawn from
the origin determine a triangle OP1P2, whose area A can easily be ex-
pressed if the lengths ri , r2 and direction cosines
of the vectors are given. For, denoting the angle
Pi OP2 by \p we have for the area A r
A = \ riTi sin ^,
where sin ^ can be expressed in terms of the direc-
tion cosines by § 301.
yQz
Fig. 129
305. Moment of a Force. Such areas are used in mechanics to
represent the moments of forces. The moment of a force about a point O
is defined as the product of the force into the
perpendicular distance of from the line of
action of the force. Thus, if the vector P1P2
(Fig. 130) represent a force (in magnitude,
direction, and sense) the 'moment of this force
about the origin is equal to twice the area
of the triangle OP1P2, i.e. to the area of the
parallelogram OP1P2P3, where OP3 is a vector
equal to the vector P1P2. Fig. 130
It is often more convenient to represent this moment not by such an
area, but by a vector OQ, drawn from O at right angles to the triangle,
and of a length equal to the number that represents the moment. If the
body on which the force acts could turn freely about this perpendicular
the moment would represent the turning effect of the force P1P2.
XIII, § 306]
COORDINATES
289
The 'sense of this vector that represents the inoiuent is taken so as to
make the vector point toward that side of the plane of the triangle from
which the force P1P2 is seen to turn counterclockwise.
306. If we square the expression found in § 304 for the area of the
triangle OP1P2 and substitute for sin'^^j/ its value from § 301, we find :
A^
= \n^r2^(^
mi
Wi
2
ni
h
2 h mi
m2
nz
+
Wo
h
■" I2 m2
1
Hence A^ is the sum of the squares of the three quantities
Ax = i riVi
mi Wi
1712 W2
A,
ni h
n.2 h
A^=\ rira
which have a simple geometrical and mechanical interpretation. For, as
the coordinates of Pi , Po are
xi = hn, yi = wiiri, zi = niVi,
Xi = hr2', y2 = m2r2, Z2 = ^2^2,
we have,
e.g.,
A.= l
hn rrnn
hrz wi2r2
= i
xi yi
X2 2/2
and as Xi , yi and X2 , 2/2 are the coordinates of the projections ^1 , ^2 of
Pi , P2 on the plane Oxy, Az represents (§ 12) the area of the triangle
0Q\Q2-, i-e. the projection on the plane Oxy of the area OP1P2. Sim-
ilarly, Aj. and Ay are the projections of the area OP1P2 on the planes
Oyz and Ozx, respectively. As any three mutually rectangular planes
can be taken as coordinate trihedrals, our formula A^ = A^ + A^ + A^
means that the square of the area of any triangle is equal to the sum of
the squares of its projections on any three mutually rectangular planes.
In mechanics, 2 A^ is the moment of the projection Qi Q2 of the force
P1P2 about 0, or what is by definition the same thing, the moment of
P1P2 about the axis Oz. Similarly, for 2^^, 2 Ay. The proposition
means, therefore, that the moments of P1P2 about the axes Ox, Oy, Oz
laid off as vectors along these axes can be regarded as the rectangular
components of the moment of P1P2 about the point ; in other words,
2 ^,, 2 Ay, 2 Ag are the components along Ox, Oy, Oz of that vector
2 ^ (§ 305) which represents the moment of P1P2 about O.
u
290
SOLID ANALYTIC GEOMETRY [XIII, § 307
307. Polar Coordinates. The position of any point P {¥\\
131) can also be assigned by its
radius vector OP=r, i.e. the dis-
tance of P from a fixed origin or
pole O, and two angles : the colati-
tude 6, i.e. the angle NOP made
by OP with a fixed axis ON, the
2)olar axis, and the longitude (f>,
i.e. the angle AOP' made by the
plane of 9 with a fixed plane
NOA through the polar axis, the
initial meridian plane.
A given radius vector r confines the point P to the sphere
of radius r about the pole 0. The angles and <^ serve to
determine the position of P on this sphere. This is done as
on the earth's surface except that instead of the latitude, which
is the angle made by the radius vector with the plane of the
equator AP', we use the colatitude or polar distance = NOP.
The quantities r, 6, and <^ are the polar or spherical coordi-
nates of P. After assuming a point as pole, a line ON
through 0, with a definite sense, as polar axis, and a (half-)
plane through this axis as initial meridian plane, every point
P has a definite radius vector r (varying from zero to infinity),
colatitude 6 (varying from to tt), and a definite longitude <^
(varying from to 2 tt). The counterclockwise sense of rotation
about the polar axis is taken as the positive sense of <^.
308. Transformation from Cartesian to Polar Coordinates-
The relations between the cartesian coordinates x, y, z and the
polar coordinates r, 6, <j> of any point P appear directly from
Fig. 132. If the axis Oz coincides with the polar axis, the
plane Oxy with the equatorial j^lane, i.e. the plane through the
XIII, § 308] COORDINATES 291
pole at right angles to the polar axis, while the plane Ozx is
taken as initial meridian plane, the pro- ^
jections of OP = r on the axis Oz and ^
on the equatorial plane are
OR = rGO^e, OQ = r sine.
Projecting OQ on the axes Ox, Oy,we
find Fig. 132
x=r sin 6 cos <^, y = r sin 6 sin <^, z = r cos 6.
Also r = Vx^-\-y^ -\- Z-, cosO = — ^ tan<^ = '^.
Va^ -h y2 ^ ;32 X
EXERCISES
1. Find the area of the triangle whose vertices are (a, 0, 0), (0, 6, 0),
(0, 0, c).
2. Find the area of the triangle whose vertices are the origin and the
points (3, 4, 7), (- 1, 2, 4).
3. Find the area of the triangle whose vertices are (4, — 3, 2),
(6,4,4), (-5, -2, 8).
4. The cartesian coordinates of a point are 1, VS, 2\/3 ; what are its
polar coordinates ?
5. If r = 5, = i TT, = ^ TT, what are the cartesian coordinates ?
6. The earth being taken as a sphere of radius 3962 miles, what are
the polar and cartesian coordinates of a point on the surface in lat. 42° 17'
N. and long. 83° 44' W. of Greenwich, the north polar axis being the axis
Oz and the initial meridian passing through Greenwich ? What is the
distance of this point from the earth's axis ?
7. Find the area of the triangle whose vertices are (0, 0, 0) , (ri, 0i, 0i),
(ra, 02, 02).
8. Express the distance between any two points in polar coordinates.
9. Find the area of any triangle when the cartesian coordinates of the
vertices are given.
10. Find the rectangular components of the moment about the origin
of the vector drawn from (1, — 2, 3) to (3, 1, — 1).
CHAPTER XIV
THE PLANE AND THE STRAIGHT LINE
PART I. THE PLANE
309. Locus of One Equation. In plane analytic geometry
any equation between the coordinates a:, y or r, <^ of a point in
general represents a plane curve. In particular, an equation of
the first degree in x and y represents a straight line (§ 30);
an equation of the second degree in x and y in general repre-
sents a conic section (§ 245).
In solid analytic geometry any equation between the coordi-
nates ic, y, z or ?', d, <^ of a point in general represents a surface.
Thus, if any equation in x, y, z,
F{x,y,z) = 0,
be imagined solved for z so as to take the form
2;=/(a;, y),
we can find from this equation to every point (a;, y) in the
plane Oa^j one or more ordinates z (which may of course be
real or imaginary), and the locus formed by the extremities of
the real ordinates will in general form a surface. It may how-
ever happen in particular cases that the locus of the equation
F(x, y, z) = 0, i.e. the totality of all those points whose coordi-
nates x, yj z when substituted in the equation satisfy it, con-
sists only of isolated points, or forms a curve, or that there are
no real points satisfying the equation.
Similar considerations apply to an equation in polar
coordinates
F(r, $,<!>) =0.
202
XIV, §311] THE PLANE • 293
310. Locus of Two Simultaneous Equations. Two simulta-
neous equations in x, y, z (or in the polar coordinates r, 6, </>)
will in general represent a curve in space, namely, the inter-
section of the two surfaces represented by the two equations
separately.
Thus, in the present chapter, we shall see that an equation of
the first degree in x, y, z represents a plane and that therefore
two such equations represent a straight line, the intersection of
the two planes. In chapters XV and XVI we shall discuss
loci represented by equations of the second degree, which are
called quadric surfaces.
311. Equation of a Plane. Every equation of the first degree
in X, y, z represents a plane. The plane is defined as a surface
such that the line joining any two of its points lies completely
in the surface. We have therefore to show that if the general
equation of the first degree
(1) Ax-{-By-\-Cz+D =
is satisfied by the coordinates of any two points Pi(x^y y^ z^
and P2fe> Vi) %)> ^•^' if
^ + ^yi + Czi + Z> = 0,
Ax^-\-By^+Cz^-\-D=^0,
then (1) is satisfied by the coordinates of every point
P{x, y, z) of the line PiP^.
Now, by § 295, the coordinates of every point of the line
P^Pi can be expressed in the form
x = x^-\-k(x^-x;), y = yi + k(y2~yi), z = Zj + A:(% - Zj),
where k is the ratio in which P divides PiP^j i-e.
k = P,P/P,P,.
We have therefore to show that
A[x^ + kix^ - a^)] + B[y^ +k{y, -yO] + C[_z,-\-k{z,-z,)-] +/)=0,
(2)
294 SOLID ANALYTIC GEOMETRY [XIV, § 311
■whatever the value of k. Adding and subtracting kD, we can
write this equation in the form
(1 - k)(Ax,-{- By^-^ Cz:, + D) -i-kiAx^-^ By,-{- Cz2-\- D) :=0',
and this is evidently true for any k, owing to the conditions (2).
312. Essential Constants. The equation (1) will still rep-
resent the same plane when multiplied by any constant differ-
ent from zero. Since A, h, C cannot all three be zero, we
can divide (1) by one of these constants ; it will then contain
not more than three arbitrary constants. We sa}^ therefore
that the general equation of a plane contains th7'ee essential
C07istants. This corresponds to the geometrical fact that a
plane can, in a variety of ways, be determined by three condi-
tions, such as the conditions of passing through three points,
etc.
313. Special Cases. If, in equation (1), D = 0, the plane
evidently passes through the origin.
If, in equation (1), 0=0, so that the equation is of the
form
Ax + By-\-D = 0,
this equation represents the plane perpendicular to the plane
Oxy and passing through the line whose equation in the
plane Oayy is xlx -i-By -\- D = 0. For, the equation Ax + By
-h Z> = is satisfied by the coordinates of all points (x, y, z)
whose x and y are connected by the relation Ax -\- By -\- D =
and whose z is arbitrary, but it is not satisfied by the coordi-
nates of any other points. Similarly, if ^ = in (1), the plane
is perpendicular to Ozx ; if ^ = 0, the plane is perpendicular to
Oyz.
If 5 = and 0=0 in (1), the equation obviously represents
a plane perpendicular to the axis Ox ; and similarly when
and A, or A and B are zero.
XIV, §315] THE PLANE 295
Notice that the line of intersection of (1) with the plane
Oxy, for instance, is represented by the simultaneous equations
Ax + By-\-Cz + D=^0,z = (i.
314. Intercept Form, li D^O the equation (1) can be
divided by Z>; it then assumes the form
D D^ D
If A, B, C are all different from zero, this equation can be
written
^ I y I ^ ^1
-D/A^ -D/B^ -D/C '
or, putting - D/A = a, - D/B = b, - D/C= c :
(3) ^ + | + ?=1.
a b c
In this equation, called the intercept form of the equation
of a plane, the constants a, b, c are the intercepts made by the
plane on the axes Ox, Oy^ Oz respectively. For, putting, for
instance, y = and z = 0, we find x = a\ etc.
315. Plane through Three Points. If the plane
Ax + By+Cz + D =
is to pass through the three points Px{x^, yi, Zi), ^2(^2? 2/2? %)>
Ps(xs, 2/3, 23), the three conditions
' Ax, + By,-\-Cz,+D=0,
Ax,^ + %2 + Cz. 4- Z) = 0,
Ax, + By,-^Cz, + D =
must be satisfied. Eliminating A, B, (J, D between the four
linear homogeneous equations (compare § 75) we find the equa-
tion of the plane passing through the three points in the form
X y z 1
^1 2/1 ^i 1
»2 2/2 2, 1
•^3 2/3 '^Z -I
= 0.
296
SOLID ANALYTIC GEOMETRY [XIV, § 315
EXERCISES
1. Find the intercepts made by the following planes :
(a) 4 a; + 12 ?/ + 3 ;2 = 12 : (b) 15 x - 6 1/ + 10 ^ + 30 = ;
(c) x-y -\-3-l=0; (d) x-^2y -\-S.z + 4 = 0,
2. Interpret the following equations :
(a) x+y + z = l; (b) 6y - ^ z = 12 ;
(c) x + y=0; (d) 6y + l2=0.
'" 3. Find the plane determined by the points (2, 1, 3), (1, —5,0),
(4,6, -1).
4. Write down the equation of the plane whose intercepts are 3, 2, — 5.
5. Find the intercepts of the plane passing through the points
(3, -1,4), (6,2,-3), (-1, -2, -3).
6. If planes are parallel to and a distance a from the coordinate planes,
what are their intercepts ? What are their equations ?
7. Show that the four points (4,3,3), (4,-3,-9), (0,0,3),
(2, 1, 2) lie in a plane and find its equation.
316. Normal Form. The position of a plane in space is
fully determined by the length p = ON (Fig. 133) of the per-
pendicular let fall from the origin
on the plane and the direction co-
sines I, m, n of this perpendicular
regarded as a vector ON. Let Pbe
any point of the plane and OQ=x,
QR = y, HP— z its coordinates ; as
the projection of the open polygon
OQRP on ON is equal to ON
(§ 294) we have
(4) Ix -\- my -\- nz = p.
This equation is called the normal form of the equation of a
plane. Observe that the number p is always positive, being
the distance of the plane from the origin, or the length of the
vector ON Hence Ix + my -\- nz is always positive.
Fig. 133
XIV, §317] THE PLANE 297
317. Reduction to the Normal Form. The equation Ax +
By -{- Cz -\- D = is in general not of the form lx+my+nz=p
since in the latter equation the coefficients of x, y, z, being the
direction cosines of a vector, have the property that the sum
of their squares is equal to 1, while A^-\-B--{- C^ is in general
not equal to 1. But the general equation can be reduced to
the normal form by multiplying it by a constant factor k
properly chosen. The equation
kAx + kBy + kCz + kD =
evidently represents the same plane as does the equation
Ax + By H- Cz + D = 0; and we can select k so that
{kAy + (kBf + (kCy = 1, viz. k= ^
±VA'-\-B^-\-C^
As in the normal form the right-hand member p is positive
(§ 316) the sign of the square root should be selected so that
kD becomes negative.
The normal fonn is therefore obtained by dividing the equation
Ax + By -\- Gz-^D = Oby ± VA^ + B^ -^ G^ according as D is
negative or positive.
It follows at the same time that the direction cosines of any
normal to the plane Ax -\- By + Cz -\- D = are proportional
to Ay Bf C, viz.
; A B
1= , m =
±V^2 + 52+ C2 ± V^2_^52_^02
±VA''-\-B'-\- (?
and that the distance of the plane from the origin is
P
± VA? -f- 52 ^ C2
the upper sign of the square root to be used when D is nega-
tive, the lower when D is positive.
298
SOLID ANALYTIC GEOMETRY [XIV, § 318
Fig. 134
318. Distance of Point from Plane. Let Ix -{- my -i-nz =p
be the equation of a plane in the normal form, Pi{xi, y^, z^
any point not on this plane (Fig. 134). The projection OS of
the vector OP^ on the normal to the
plane being equal to the sum of the
projections of its components 0Q =
Xij QR = 2/i, RP\ = 2;i, we have
OS = lx^-\- myi + nzi.
Hence the distance d of Pi from the
plane, which is equal to NS, will be
d = OS — 0N= Ix^ + myi + nzy^ — p.
If this expression is negative, the point P^ lies on the same
side of the plane as does the origin ; if it is positive, the point
Pi lies on the opposite side of the plane. Any plane thus di-
vides space into two regions, in one of which the distance of
every point from the plane is positive, while in the other the
distance is negative. If the plane does not pass through the
origin, the region containing the origin is the negative region ;
if it does, either side can be taken as the positive side.
To find the distance of a point Pi(.'«i, y^ Zi) from a plane
given in the general form
Ax-\-By + Cz-{-D = 0,
we have only to reduce the equation to the normal form
(§ 317) and then to substitute for x, y, z the coordinates Xi, yi,
Zi of Pi; thus
^ ^ ^Xi + By, + Cz^^- D
the square root being taken with + or — according as D is
negative or positive.
Notice that d is the distance from the plane to the point
Pi , not from Pi to the plane.
XIV, § 320] THE PLANE 299
319. Angle between Two Planes. As two intersecting
planes make two angles whose sum = tt, we shall, to avoid any
ambiguity, define the angle between the planes as the angle
between the perpendiculars (regarded as vectors) drawn from
the origin to the two planes.
If the equations of the planes are given in the normal form,
kx + m^ + n^z = pi,
l^ + m^ + n.^ = 2)2,
we have, by § 299, for the angle if/ between the planes :
cos \p = IJ2 + wi^ma + ri^ng.
If the equations of the planes are in the general form,
A^ + B^ + C2Z + A = 0,
we find by reducing to the normal form (§ 317) :
cos l{/ =
A,A,-\-B,B2+C,C,
± VA' + B,' + 0^2 . ± VA^ + ^^2 + c^2
320. Bisecting Planes. To find the equations of the two
planes that bisect the angles formed by two intersecting planes
given in the normal form,
liX + Miy + n^z —pi = 0, l^-\- m<^ + n2Z — pg = 0,
observe that for any point in either bisecting plane its distances
from the two given planes must be equal in absolute value.
Hence the equations of the required planes are
l^x + m^y -\-n^z—p^ = ± Q^ + m^ + n^z — ^2)-
To distinguish the two planes, observe that for the plane that
bisects that pair of vertical angles which contains the origin
the perpendicular distances are in the one angle both positive,
in the other both negative; hence the plus sign gives this
bisecting plane.
300 SOLID ANALYTIC GEOMETRY [XIV, § 320
If the equations of the planes are given in the general form,
first reduce the equations to the normal form (§ 317).
EXERCISES
1. A line is drawn from the origin perpendicular to the plane
a; — ?/ — 50 — 10 = 0; what are the direction cosines of this line ?
2. Find the distance from the origin to the plane 2x + 2?/ — = 6.
3. Find the distances of the following planes from the origin :
(a)3a'-4y + 50-8 = O, {h) x + y+ z = {),
(^c) 2y -^z = S, (d) 3x-^y + 6 = 0.
4. Find the distances from the following planes to the point
(2, 1, - 3) :
(a) 3 X + 5 1/ - 6 = 8, (&) 2x-Sy - z = 0, (c) x + y + z=0.
5. Find the plane through the point (4, 8, 1) which is perpendicular
to the radius vector of this point ; also the parallel plane whose distance
from the origin is 10 and in the same sense.
6. Find the plane through the point (— 1, 2, — 4) that is parallel to
the plane ix — Sy + 2z = 8; what is the distance between these planes ?
7. Find the distance between the planes 4x — 5?/ — 2^ = 6, 4lX — 6y
-20 + 8 = 0.
8. Are the points (6, 1, — 4) and (4, — 2, 3) on the same side of the
plane 2x + Sy - 5z + 1 =0?
9. Write down the equation of the plane equally inclined to the axes
and at the distance p from the origin.
10. Show that the relation between the distance p from the origin to a
plane and the intercepts a, &, c is 1/a^ + 1/b^ 4- 1/c^ = 1/p^.
11. Show that the locus of the points equally distant from the points
Pi(xi, ?/i, 0i) and Po^x-y, 2/2, ^2) is a plane that bisects P1P2 at right
angles.
12. Find the equations of the planes bisecting the angles: (a) between
the planes x + y -{- z-3=0, 2x-3i/H- 4^ + 3 = 0; (6) between the
planes 2x — 2y — z = 8,x-\-2y — 2z = 6.
XIV, § 321]
THE PLANE
301
321. Volume of a Tetrahedron. The volume of the tetrahe-
dron whose vertices are the points Pi(x^, yi, Zy), Pz^x^, 2/2? ^2)^
^(•^*3) 2/3) h)) Pii^if 2/4) ^a) can be expressed in terms of the
coordinates of the points. The equation of the plane deter-
mined by the points P2 , P3 , P4 is (§ 315)
X y z 1
^2 2/2 Z2 1
•^3 y^ ^3 1
^4 yi z, 1
Now the altitude d of the tetrahedron is the distance from this
plane to the point P^ (x^ , y^ , z^), i.e. (§ 318)
2/1 ^i 1
= 0.
2/2 ^2 1
2
2^2 ^2 1
2
«2 2/2 1
yz Zz 1
+
2:3 X3 1
+
a^3 2/3 1
2/4 2:4 1
Z, X, 1
^4 2/4 1
(^=
But the denominator is seen immediately to represent twice
the area of the triangle with vertices P^, P^, P^ (Ex. 9, p. 291),
i.e. twice the base of the tetrahedron. Denoting the base by jB,
we then have
Xl 2/1 Zy 1
^'2 2/2 ^2 1
^Z 2/3 2!3 1
X, 2/4 2:4 1
The volume of the tetrahedron is V= ^Bd, and therefore
Xy 2/1 ^1 1
2 2/2 '^2 -^
•^3 2/3 2!3 1
X, 2/4 ^4 1
2Bd
302
SOLID ANALYTIC GEOMETRY [XIV, § 322
322. Simultaneous Linear Equations. Two simultaneous
equations of the first degree,
A^ + B^y + (7,2 -h A = 0,
represent in general the line of intersection of the two planes
represented by the two equations separately. For, the coordi-
nates of every point of this line, and those of no other point,
satisfy both equations. See § 310 and §§ 326-327.
Three simultaneous equations of the first degree,
A,x-{-B,y-\-G,z + D, = 0,
A^ + B^-\- C.Z + A = 0,
A,x + B,y + Csz + A = 0,
determine in general the point of intersection of the three
planes. The coordinates of this point are found by solving
the three equations for x, y, z. But it may happen that the
three planes have no common point, as when the three lines of
intersection are parallel, or when the three planes are parallel ;
and it may happen that the planes have an infinite number of
points in common, as when two of the planes, or all three,
coincide, or when the three planes pass through one and the
same line.
Four planes will in general have no point in common. If they do, i.e.
if there exists a point (xi , yi , zi) satisfying the four equations
Aixi + Bm + C\zi + i)i = 0,
A2X1 + BiVi + 02^1 + Z)2 = 0,
Asxi + Bm + GzZi + Z>3 = 0,
A^Xi + BaVi + C4Z1 + 2)4 = 0,
1 between these equations so that we find
= 0.
can eliminate xi , yi,
^u
1 between th
condition
A,
Bi Ci Di
A2
B2 C2 D2
As
Bs Cs D,
A,
B, C4 2>4
XIV, §323] THE PLANE 303
EXERCISES
1. Find the volume of the tetrahedron whose vertices are (0, 0, 0),
(a, 0,0), (0, &, 0), (0, 0, c).
2. Find the volumes of the tetrahedra whose vertices are the following
points :
(a) (7, 0, 6), (3, 2, 1), (- 1, 0, 4), (3, 0, - 2).
(6) (3, 0, 1), (0, - 8, 2), (4, 2, 0), (0, 0, 10).
(c) (2, 1, - 3), (4, - 2, 1), (3, -7, - 4), (5, 1, 8).
3. Find the coordinates of the points in which the following planes
intersect :
(a) 2x + 67J -^ z ~2 = 0, x + 6y + z = 0, 3x— 3?/ + 2^— 12=0.
(6) 2x+y+z=a + b + c, ix—2 y-^z=2 a-2b + c, 6x~y=Sa-b.
"^ 4. Show that the four planes 6x — 3y — z = 0, 4:X — 2y-\-z = S,
Sx + 2y — 6z = 6, x -{- y + z = 6 pass through the same point. What
are the coordinates of this point ?
~ - 5. Show that the four planes 4:X + y-\-z + 4: = 0, x-\-2y — z + S=0,
y— 6z + li = 0, x + y + z — 2 = have a common point.
6. Show that the locus of a point the sura of whose distances from
any number of fixed planes is constant is a plane.
323. Pencil of Planes. All the planes that pass through
one and the same line are said to form a pencil of planes, and
their common line is called the axis of the pencil.
If the equations of any two non-parallel planes are given,
say
A,x + B,y + C,z -h A = 0,
A2X + ^22/ + O2Z 4- A = 0,
then the equation of any other plane of the pencil having their
intersection as axis can be written in the form
(2) {A,x + B,y + G,z + A) + A:(^2aj + A2/ + O^z + A) = 0,
where /c is a constant whose value determines the position of
the plane in the pencil.
For, this equation (2) being of the first degree in x, y, z
certainly represents a plane ; and the coordinates of the points
304 SOLID ANALYTIC GEOMETRY [XIV, § 323
of the line of intersection of the two given planes (1), since
they satisfy each of the equations (1), must satisfy the equa-
tion (2) so that the plane (2) passes through the axis of the
pencil.
324. Sheaf of Planes. All the planes that pass through
one and the same point are said to form a sheaf of planes, and
their common point is called the center of the sheaf.
If the equations of any three planes, not of the same pencil,
are given, say
A^x + ^22/+ 022; + A = 0,
A,x-\-B,y-\-C,z-\-D, = 0,
then the equation of any other plane of the sheaf having their
point of intersection as center can be written in the form
(A,x + B,y + C,z 4- A) + ^i (A,x + B,y + C,z + A)
+ k, {A,x + B^ 4- C,z + A) = 0,
where ki and k2 are constants whose values determine the
position of the plane in the sheaf.
The proof is similar to that of § 323.
*
325. Non-linear Equations Representing Several Planes.
When two planes are given, say
A,x + B,y + C,z-^Di = 0,
A^ + B^ + Cz + D^^O,
then the equation
(A,x + B,y -f C,z -f- D,)(A,x + B,y + 0,z -f A) = 0,
obtained by equating to zero the product of the left-hand mem-
bers (the right-hand members being reduced to zero), is satis-
fied by all the points of the first given plane as well as all the
points of the second given plane, and by no other points.
The product equation is therefore said to represent the two
given planes. The equation is of the second degree.
XIV, §325] THE PLANE 305
Similarly, by equating to zero the product of the left-hand
members of the equations of three or more planes (the right-
hand members being zero) we obtain a single equation repre-
senting all these planes. An equation of the nth degree may,
therefore, represent n planes ; it will do so if its left-hand mem-
ber can be resolved into n linear factors with real coefficients.
EXERCISES
1. Find the plane that passes through the line of intersection of the
planes 5x — 32/4-4^ — 35=0, x + y — z -.-O and through (4, — 3, 2) .
— 2. Show that the planes 3x — 22/ + 5 + 2=0, ic + ?/ — — 5 = 0,
6a; + 2/ + 20— 13 = belong to the same pencil.
3. Show that the following planes belong to the same sheaf and find
the coordinates of the center of the sheaf : 6a; + y — 42r = 0, x + |/ + = 5,
2x — 4:y-z = 10,2x + Sy+z = 4.
4. What planes are represented by the following equations ?
(a) a;2-6x + 8 = 0, (&) ^2_9 = o, (c) x'^ - z^ = 0, {d) x'^-4xy = 0.
5. Find the cosine of the angle between the following pairs of planes :
(a) 4:X-Sy-z=6, x-\-y~z=8 ; (6) 2x4-7 ^+4^=2, x-9y-2 0=12.
6. Show that the following pairs of planes are either parallel or
perpendicular :
(a) Sx-2y + 6z=0,2x+Sy=S; (b) 6x+2y-z=6, lOx+iy-2 e=S;
(c) x + y-2z = S,x+y-\-z = ll; (d) x- 2y - z = S, Sx -6y-S z=6.
7. Find the plane that is perpendicular to the segment joining the
points (3, — 4, 6) and (2, 1, — 3) at its midpoint.
8. Show that the planes Aix + Biy + Ciz + Di=0, Aix + B^y + C^z
-f-jD2 = are parallel (on the same or opposite sides of the origin) if
AxA2 + BxB2+CiC2 ' _^^
VAi^ + J?i2 + 0/ VA2^ + B2^ + CV
9. A cube whose edges have the length a is referred to a coordinate
trihedral, the origin being taken at the center of a face and the axes par-
allel to the edges of the cube. Find the equations of the faces.
306
SOLID ANALYTIC GEOMETRY [XIV, § 325
X
y
z
1
Xx
2/1
Zx
1
X2
2/2
02
1
A
B
G
10. Show that the plane through the points Pi(xi, yi , z{) and
^1 (Xi, 1/2, Z2) and perpendicular to the plane Ax -^ By -\- Cz + D =
can be represented by the equation
= 0.
11. Find those planes of the pencil 4a; — 3^ + 5^ = 8, 2x + Sy — z = 4:
which are perpendicular to the coordinate planes.
12. Find the plane that is perpendicular to the plane 2x + Sy — z = l
and passes through the points (1, 1, — 1), (3, 4, 2).
13. Find the plane that is perpendicular to the planes 4a; — 3^ + = 6,
2aj + 3?/ — 5^=4 and passes through the point (4, —1,5).
14. Show that the conditions that three planes AiX + Biy + Ciz + Z)i =0,
A2X + B2y + C2Z + i>2 == 0, AzX + Bay + C^z + D3 = belong to the same
pencil, are
Ai+k A2 _ Bi + kB2 _ Ci+kC2 _ D\ + k D2 .
As B3 O3 2>3
or, putting these fractions equal to s and eliminating k and s,
B,
Ci
2>i
B2
O2
D2
=
B,
O3
2>3
D2
2>3
Ax Di Ax Bx Ax Bx Cx
A2 = D2 A2 B2 = A2 B2 O2 =0.
A3 Dz A3 B3 A3 B3 C3
(Verify Ex. 2 by using these conditions. )
15. Find the equations of the faces of a right pyramid, with square
base of side 2 a and with altitude h, the origin being taken at the center
of the base, the axis Oz through the opposite vertex and the axes Ox, Oy
parallel to the sides of the base.
16. Homogeneous substances passing from a liquid to a solid state tend
to form crystals ; e.g. an ideal specimen of ammonium alum has the form
of a regular octahedron. Find the equations of the faces of such a crystal
of edge a if the origin is taken at the center and the axes through the
vertices, and determine the angle between two faces.
17. Find the angles between the lateral faces of a right pyramid whose
base is a regular hexagon of side a and whose altitude is h.
XIV, § 327]
THE STRAIGHT LINE
307
PART II. THE STRAIGHT LINE
326. Determination of Direction Cosines. Two simulta-
neous linear equations (§ 322),
(1) AiJC + By+Cz-\-I)=0, A'3c+B'ij+C'z-{-I)'=0,
represent a line, namely, the intersection of the two planes
represented by the two equations separately, provided the two
planes are not parallel.
To obtain the direction cosines I, m, n of this line observe
that the line, since it lies in each of the two planes, is perpen-
dicular to the normal of each plane. Now, by § 317 the direc-
tion cosines of these normals are proportional to A, B, C and
A', B', C, respectively. We have therefore
Al + Bm-j-Cn = 0, A'l -j- B'm -{- C'n = 0,
whence
l:m:7i =
BC
B'C'l
CA
C'A'
AB
A'B'
The direction cosines themselves are then found by dividing
each of these determinants by the square root of the sum of
their squares.
327. Intersecting Lines. The two lines
A^x -f B,y + C,z + A = 0, ]
A,'x-\-B,'y-rC,'z-}-D,'=0 J
I A^'x+B^'y^ C^z-^D^ =
will intersect if, and only if, the four planes represented by
these equations have a common point. By § 322, the condition
for this is
A^ B, C, A
A,' B,' C A
A2 X>2 C2 -U2
A^ Bi C^ A' "'I'
= 0.
308 SOLID ANALYTIC GEOMETRY [XIV, § 328
328. Special Forms of Equations. For many purposes it is
convenient to represent a line by means of one of its points
and its direction cosines, or by means of two of its points.
Let the line be called X.
If (^a, 2/i> %) is a given point of A. and I, m, n are the direc-
tion cosines of A, then every point (x, y, z) of A. must satisfy
the relations (§ 298) :
^ ^ I m n '
In these equations, Z, m, n, can evidently he replaced by any
three numbers proportional to I, m, n. Thus, if (^2, y2, z.,) be
any point of A, different from (a^, 2/1, ^j), we have the continued
proportion
0^2 — ajj : 2/2 — 2/1 : 2!2 — 2i = Z : m : n ;
hence the equations of the line through the two points (x^ , 2/1 j ^1)
and (X2 , 2/2 , Z2) are :
(3) 00-iCt ^ y-Vi ^ g-^i ^
i»2-«i 2/2-2/1 «2-«i*
If, for the sake of brevity, we put x^— x^ = a, 2/2 — 2/i = ^>
Z2 — Zi=: c, we can write the equations of the line in the form
^ a b c '
where a, b, c, are proportional to I, m, n, and can be regarded as
the components of a vector parallel to the line.
The equations (3) also follow directly by eliminating k be-
tween the equations of § 295, namely,
(5) a?=a?i-f-A;(a?2-a5i), y=y^^k{y^-y{), z=z^-]-Jc(z^-z{).
These equations which, with a variable h, represent any point
of the line through (iCi , 2/1 , ^j) and {x^ , 2/2 > ^2) are called the
parameter equations of the line.
XIV, § 329]
THE STRAIGHT LINE
309
329. Projecting Planes of a Line. Each of the forms (2),
(3), (4), which are not essentially different, furnishes three
linear equations ; thus (4) gives :
y
c
a
Vi
he c a ah
but these three equations are equivalent to only two, since from
any two the third follows immediately.
The first of these equations, which
can be written in the form
cy-hz-{cyy-hz;)=0,
represents, since it does not contain x
(§ 313), a plane perpendicular to the
plane 0?/2; and as this plane must con-
tain the line X it is the plane CCA
that projects \ on the plane Oyz (Fig. 135). Similarly the other
two equations represent the planes that project \ on the co-
ordinate planes Ozx and Oxy. Any two of these equations
represent the line X as the intersection of two of these pro-
jecting planes.
At the same time the equation
Fig. 135
can be interpreted as representing a line in the plane Oyz,
viz. the intersection of the projecting plane with the plane
£C = 0. This line {AC in Fig. 135) is the projection X^ of X on
the plane Oyz. As the other two equations (4) can be inter-
preted similarly it appears that the equations (2), (3), or (4)
represent the line X by means of its projections A^., X^, A, on
the three coordinate planes, just as is done in descriptive
geometry. Any two of the projections are of course sufficient
to determine the line.
310 SOLID ANALYTIC GEOMETRY [XIV, § 330
330. Determination of Projecting Planes. To reduce the
equations of a line A given in the form (1) to the form (4) we
have only to eliminate between the equations (1) first one of
the variables x, y, z, then another, so as to obtain two equa-
tions, each in only two variables (not the same in both).
The process will best be understood from an example. The
line being given as the intersection of the planes
(a) 2x-3y-\-z + 3=:0,
{b) x-\-y-j-z-2 = 0,
eliminate z by subtracting (b) from (a) and eliminate x by
subtracting (6), multiplied by 2, from (a) ; this gives the line
as the intersection of the planes
x — 4:y -\-5 = 0j
-5y-^z + 7 = 0,
which are the projecting planes parallel to Oz and Ox, i.e. the
planes that project the line on Oxy and Oyz. Solving for y
and equating the two values of y we find :
x-\-5 _y _ z — T
4 "1"" -5*
The line passes therefore through the point (—5, 0, 7) and
has direction cosines proportional to 4, 1, — 5, viz.
,4 1 5
I — , m
V42 V42 V42
EXERCISES
1. Write the equations of the line through the point (— 3, 1, 6) whose
direction cosines are proportional to 3, 5, 7.
2. Write the equations of the line through the point (3, 2 — 4) whose
direction cosines are proportional to 5, — 1, 3.
3. Find the line through the point (a, 6, c) that is equally inclined
to the axes of coordinates.
XIV, § 331] THE STRAIGHT LINE 311
4. Find the lines that pass through the following pairs of points :
. (a) (4, - 3, 1), (2, 3, 2), (&) (- 1, 2, 3), (8, 7, 1),
(c) (-2,3, -4), (0,2,0), (d) (-1, -5, -2),, (-3,0,-1),
and determine the direction cosines of each of these lines.
5. Find the traces of the plane 2 x — 3 «/ - 4 ^ = 6 in the coordinate
planes.
^ 6. Write the equations of the\me2x-Sy + 5 z~6=^0,x—y+2z-S=0
in the form (4) and determine the direction cosines.
7. Put the line 4:X — Sy — 6 = 0, x-y-z-4: = in the form (4)
and determine the direction cosines.
8. Find the line through the point (2, 1,-3) that is parallel to the
]ine2x-Sy-\-4z-6 = 0, 6x + y-2z-S = 0.
9. What are the projections of the line 5x — 3?/ — 7^; — 10 = 0,
X -\-y — S z +6 = on the coordinate planes ?
10. Obtain the equations of the line through two given points by-
equating the values of k obtained from § 295.
11. By § 317, the direction cosines of any line are proportional to the
coefficients of x, y, and z in the equation of a plane perpendicular to the
line. Find a line through the point (3, 5, 8) that is perpendicular to the
plane 2x + i/ + 30 = 5.
331. Angle between Two Lines. The cosine of the angle ^p be-
tween two lines whose direction cosines are h, mi, wi and h, rrn, nz is,
by § 299,
cos \p = hh + W2im2 + wiW2.
Hence if the lines are given in the form (4) , say
x-x i _ y — y i _ z - zi x-xi _ y — yt _ z — zt
«i &i c'l ai 62 C2
we have
cos V = . ^^^^ "^ ^^^2 "•" ^^^2
± Vai2 + &i2 + ci2 . ± Va22 + 62^ + C22
If the lines are parallel, then
ai_ &i _ci.
a2, 02 C2
if they are perpendicular, then
aia2 + 61&2 + C1C2 = ;
and vice versa*
312
SOLID ANALYTIC GEOMETRY [XIV, § 332
Let the line and plane
332. Angle between Line and Plane.
be given by the equations
x — xi _ y- y\ _ z — zi
a h c '
Ax-\- By + Cz-\r D = (i.
The plane of Fig. 136 represents the plane
through the given line perpendicular to the given
plane. The angle /3 between the given line and
plane is the complement of the angle a between the line and any perpen-
dicular PiV to the plane. Hence
.„«_ aA-^hB-\-cC
Fig. 136
± v/a2 + 6-2 + C2 . ± V^2 + ^2 + (72
The (necessary and sufficient) condition for parallelism of line and
plane is
aA + hB + cC = 0\
the condition of perpendicularity is
a_ _h _ c^
A~ B~ G
333. Line and Plane Perpendicular at Given Point. If the
plana Ax + By -{■ Cz -{■ D = Q
passes through the point Pi(xi , yi , zi), we must have
Axi + Byx + Czx + D = ^.
Subtracting from the preceding equation, we have as the equation of
any plane through the point Pi(xi, yi, z\) :
A(x - xi) + Biy - yx) + C{z - zi) = 0.
The equations of any line through the same point are
x — xi _ y -yi _ z — zi ^
a b c
If this line is perpendicular to the plane, we must have (§ 332) : a/ A =
b/B = c/C. Hence the equations
x — xi_y—yi_z — zi
represent the line through Pi(iCi, yi, zi) perpendicular to the plane
Aix - xi) + Biy - 2/0 + C{z - zi) = 0.
XIV, § 335]
THE STRAIGHT LINE
313
If the equations of
Fig. 137
334. Distance of a Point from a Line.
the line X are given in the form
x—xi _ y — yi _ z — zi "
I m n
where {xi , yi , z\) is a point Pi of X (Fig.
187), the distance d = QP2 of the point
•P2(a;2, ^2, Z2) from X can be found from
the right-angled triangle Pi QP2 which gives
cP = FiP2^ - PiQ^,
by observing that
P1P22 = (X2 - XiY + (2/2 -yi)2 + (Z2 - Zi)\
while PiQ is the projection of P1P2 on X. This projection is found
(§ 294) as the sum of the projections of the components X2 — xu y'z — yu
Z2 — z\ of P1P2 on X :
PiQ = 1(X2 — xi) + m(y2 - y\) + n{z2 - z{).
Hence
(?2=:(a;2-xi)2+ (y2-2/i)2 + (^2-^i)2-[Z(a;2-xi)+m(?/2-yi) +n{z2-z{)Y'
335. Shortest Distance between Two Lines. Two lines
Xi , X2 whose equations are given in the form
- 2/1 _ Z—Zi X-X2 _ y — y2 _ Z- Z2
X — X\
h
mi
W2
W2
will intersect if their directions {h , mi , ni), (I2-, wi2, n2), and the direc-
tion of the line joining the points {x\ , yi , z{)^ (x.2 , 2/2 , ^2) are complanar
(§302), i.e. if
X2 — Xi 2/2 - 2/1 Z2 - Z\
l\ m\ n\
I2 WI2 W2
If the lines Xi , X2 do not intersect, their shortest distance d is the dis-
tance of P2(aj2, ^2, Z2) from the plane through Xi parallel to X2. As this
plane contains the directions of Xi and X2 , the direction cosines of its nor-
mal are (§ 301) proportional to
mi
ni
ni h
h mi
m2
«2
»
n2 h
5
I2 m2
314
SOLID ANALYTIC GEOMETRY [XIV, § 335
and as it passes through Pi {xi , yi , ^i) its equation can be written in the
form
x — xi y — yiz — zi
h wii wi = 0.
h Wl2 W2
Hence the shortest distance of the lines Xi, X2 is :
d =
V
X2 -xi y2- yi
Z2-
-^1
h Wli Wi
h m2 W2
n
n
il Wi
l2 W2
2
+
ni Zi
W2 h
2
+
h
h
mi
m2
As the denominator of this expression is equal to sirn/' (§ 301), we have
X2 — Xi ?/2 — 2/1 2;2 - ^1
d sin xp = li mi ui
I2 Wl2 W2
EXERCISES
— '" 1. Find the cosine of the anojle between the lines
X
^-yj:zA-^±l^ ^
l_ y -3 _ g + 3
-1 2 3
2 3 4
2. Find the angle between the lines 3x — 2?/ + 42 — 1 = 0,
2x + y— 3^ + 10 = 0, and x-\-y -\- z = Q, 2x + 3y-5;s = 8.
3. Find the angle between the lines that pass through the points
(4, 2, 5), (- 2, 4, 3) and (- 1, 4, 2), (4, - 2, - 6).
4. Find the angle between the line
a; + l _ y-2 _ g+ 10
3 -5 3
and a perpendicular to the plane 4a: — 3?/ — 2^ = 8.
6. In what ratio does the plane 3x— 4i/ + 65; — 8 = divide the
segment drawn from the origin to the point (10, — 8, 4).
6. Find the plane through the point (2, — 1, 3) perpendicular to the
line
x — '6 y + 2 _z — 7
XIV, §335] THE STRAIGHT LINE 315
7. Find the plane that is perpendicular to the line ^x-{-y — z=6,
3x + 4?/ + 82+ 10 =0 and passes through the point (4, —1,3),
*— 8. Find the plane through the origin perpendicular to the line
Sx-2y + z=6, Sx + y -4z = S.
9. Find the plane through the point (4, — 3, 1) perpendicular to the
line joining the points (3, 1, — 6), (— 2, 4, 7).
10. Find the line through the point (2, — 1, 4) perpendicular to the
plane x — 2y-\-4:Z = 6.
■"" 11. Show that the lines x/S = y/ —1 = z/—2 and x/4: — y/6 = z/S are
perpendicular.
— 12. Show that the lines
^izi = L+2^£j::^ ^^^ x-2^y-S^ z
1-2 3 _2 4 -6
are parallel.
— 13. Find the angle between the line 3 ic — 2 y — ^ = 4, 4 a; + 3 ?/ — 3 ^ = 6
and the plane x -\-y -\- z — %.
14. Find the lines bisecting the angles between the lines
X— O' - V - h _ z — c ^^^^ x — a __ y— h _ z — c
h nil ni h m^ nt
15. Find the plane perpendicular to the plane Zx — ^y — z — Q and
passing through the points (1, 3, — 2), (2, 1, 4).
16. Find the plane through the point (3, — 1, 2) perpendicular to the
line 2x — 3y — 42; = 7, x-\- y — 2z = ^.
17. Find the plane through the point (a, 6, c) perpendicular to the
line Axx + Bxy + Cxz + Z)i = 0, Aix + Bty + Ciz + Z>2 = 0.
18. Find the projection of the vector from (3, 4, 5) to (2, — 1, 4) on the
line that makes equal angles with the axes ; and on the plane
2x-3?/ + 4^=6.
19. Find the distances from the following lines to the points indicated :
^""^ 1 = ^ = 4^' (0,0,0);
(6) 2x + y-5r = 6, a:-?/ + 4^ = 8, (3, 1,4);
(c) 2a; + 32/ + 50 = l, 3x-6?/ + 3;?=0, (4, 1, -2).
316 SOLID ANALYTIC GEOMETRY [XIV, § 335
20. Show that the equation of the plane determined by the line
x — xi _ y — yi _ z — zi
a b c
and the point P2 (xz , 1/2 , ^2) can be written in the form
0.
X -xi y —yi z -z\
xi -x\ yi — 7/1 zi — zi
a b c
21. Find the plane determined by the intersecting lines
x-3^y-6^ z + l and'^~^ = y~^ = ^ + ^
4 3 2 1 2 3 *
22. Find the plane determined by the line
x-xi _ y — yi _ z ~ zi
a b c ^
and its parallel through the point P2 (X2 , 2/2 , ^2).
23. Given two non-intersecting lines
x — xi _ y — yi _ z - z\ x — xi _ y -yi _ z — zi .
a\ b\ c\ at 62 C2
find the plane passing through the first line and a parallel to the second ;
and the plane passing through the second line and a parallel to the first.
24. What is the condition that the two lines of Ex. 23 intersect ?
25. Find the distance from the diagonal of a cube to a vertex not on
the diagonal.
26. Find the distance between the lines given in Ex. 23.
27. Show that the locus of the points whose distances from two fixed
planes are in constant ratio is a plane.
28. Show that the plane (w — n)x + (n — Z)?/ + (Z — m)z = contains
the line x/l = y/m = z/n and is perpendicular to the plane determined by
the lines x/m = y/n = z/l and x/n = y/l = z/m.
CHAPTER XV
^ THE SPHERE
336. Spheres. A sphere is defined as the locus of all those
points that have the same distance from a fixed point.
Let CQi, j, k) denote the center, and ?• the radius, of a sphere ;
the necessary and sufficient condition that any point F(x, y, z)
has the distance r from C{1i,j, k) is
(1) {x - hY + {y -jY -viz- ky = rK
This then is the cartesian equation of the sphere of center
C(h, j, k) and radius r.
If the center of the sphere lies in the plane Oxy^ the equa-
tion becomes
(x-hy-\-{y-jy + z'=r\
If the center lies on the axis Ox, the equation is
(x-hf+y^-\-z'^ = r\
The equation of a sphere about the origin as center is :
x2 + i/2 + s2 = r2.
337. Expanded Form. Expanding the squares in the equa-
tion (1), we find the equation of the sphere in the form
3(? + y^-\-z^-2hx-2jy-2kz-\-h''-^f-{-¥-r''=:0.
This is an equation of the second degree in x, y, z; but it is of
a particular form.
The general equation of the second degree in x, y, z is
Ax^ + By^ -\-Cz^ + 2Dyz-{-2Ezx-\-2 Fxy
+ 2Gx-^2Hy-\-2Iz-^J=0;
317
318 SOLID ANALYTIC GEOMETRY [XV, § 337
i.e. it contains a constant term J-, three terras of the first
degree, one in x, one in y, and one in z ; and six terms of the
second degree, one each in x^, y^, z"^, yz, zx, and xy.
If in the general equation we have
D = E = F=0, A = B=C=itO,
it reduces, upon division by A, to the form
x^ + f + z^ + ^x + ^-^y + ?^z + ^ = 0,
which agrees with the above form of the equation of a sphere,
apart from the notation for the coefficients.
338. Determination of Center and Radius. To determine
the locus represented by the equation
(2) Ax-'-\- Ay^ -^ Az^ + 2 Gx-\-2 By -{-2 Iz + J=:0,
where A, G, Hy 7, J, are any real numbers while ^ ^^ 0, we
divide by A and complete the squares in x, y, z\ this gives
(-!J-('-3"+('+3)"
The left side represents the square of the distance of the point
(x, y, z) from the point (— Gj A^ — H/A, — I/A) ; the right
side is constant. Hence, if the right side is positive, the equa-
tion represents the sphere whose center has the coordinates
and whose radius is
r^-^G'^^H^ + P-AJ.
A
If, however, G"^ -\- A"^ + I^ < A J, the equation is not satisfied by
any point with real coordinates. If G^ -\- H^ -{• I'^ = AJ, the
equation is satisfied only by the coordinates of the point
{-G/A,-H/A,-I/A).
XV, § 340]
THE SPHERE
319
n
Thus the equation of the second degree
Ax^ + By"^ -f- C22 + 2 Dyz + 2 Ezx + 2 Fxy
+ 2Gx^2Hy-{-2Iz-\-J=0,
represents a sphere if, and orily if
A=:B=C^O, D = E = F=0, G^-\-H^-\-P>AJ.
339. Essential Constants. The equation (1) of the sphere
contains four constants : h, j, k, r. The equation (2) contains
five constants of which, however, only four are essential since
we can divide out by one of these constants. Thus dividing
by A and putting 2 G/A = a, 2 H/A = h, 2 I/A = c, J/ A = d,
the general equation (2) assumes the form
x"^ + y^ + z"^ -\- ax -{- by -^ cz + d = Oy
with only the four essential constants a, b, c, d.
This fact corresponds to the possibility of determining a
sphere geometrically, in a variety of ways, by four conditions.
340. Sphere through Four Points. To find the equation of the
sphere passing through four points Fi(zi, ?/i, ^i), P2(X2, y^-, ^2),
^3(353, ^3, ^3), P^(Xi, y4, Z4), observe that the coordinates of these points
• must satisfy the equation of the sphere
^2 _|_ ^2 ^ ^2 ^(j^x +by +CZ + d = 0;
i.e. we must have
xi^ + Vi^ + z^ + axx + byx + csri + d = 0,
X'^ + y<^ + zci^ + ax2 + byi -^czi-^-d-^,
x-^ 4- yz^ + z^^ + axi + hyz + C23 + (^ = 0,
x^ + y^ + z^ + ax4 + by 4 -\- cz^^-d-^.
As these five equations are linear and homogeneous in 1, a, &, c, (?, we
can eliminate these five quantities by placing the determinant of their
coefficients equal to zero. Hence the equation of the desired sphere is
aj2 ^ y2 _J. ^2 a; y ^ ■
x^^y^^z^ xx y\ zi 1
X2^ + y2^ + Z2^ X2 y2 ^2 1
Xz^ + yz^ + zz^ xz yz zz 1
X4^ + 2/4^ + Z4^ Xa 2/4 Z4 1
320 SOLID ANALYTIC GEOMETRY [XV, § 340
EXERCISES
1. Find the spheres with the following points as centers and with the
indicated radii :
"^ (a) (4, -1,2), 4; (6) (0,0, 4), 4; (c) (2,-2, 1), 3; (d) (3, 4, 1), 7.
2. Find the following spheres :
- (a) with the points (4, 2, 1) and (3, — 7, 4) as ends of a diameter ;
— (6) tangent to the coordinate planes and of radius a ;
(c) with center at the point (4, 1, 5) and passing through (8, 3, — 5).
3. Find the centers and the radii of the following spheres :
(a) a;2 + ?/2 + ;s2 _ 3 X + 5 y - 6 ^ + 2 = 0.
- (6) a:2 + 2/2 + ^2 _ 2 6a; + 2 cs - &2 _ c2 = 0.
(c) 2 x2 + 2 y2 + 2 ^2 ^ 3 X - y + 5 - 11 = 0.
(d) x^ + y^-\- z^-x-y - z = 0.
__ 4. Show that the equation A{x^ -i- y^ + z^) +2 Gx + 2 Hy -\- 2 Iz +J
= 0, in which J is variable, represents a family of concentric spheres.
5. Find the spheres that pass through the following points :
— (a) (1, 1, 1), (3, - 1, 4), (- 1, 2, 1), (0, 1, 0).
(6) (0, 0, 0), (a, 0, 0), (0, 6, 0), (0, 0, c).
(c) (0, 0, 0), (- 1, 1, 0), (1, 0, 2), (0, 1, - 1).
(d) (0,0, 0), (0,0,4), (3,3,3), (0,4,0).
6. Find the center and radius of the sphere that is the locus of the
points three times as far from the point (a, 6, c) as from the origin.
— 7. Show that the locus of the points, the ratio of whose distances from
two given points is constant, is a sphere except when the ratio is unity.
— 8. Find the positions of the following points relative to the sphere
jc2 + ?/2 + ;s2_4a; + 4y-2;s = 0; (a) the origin, (6) (2, -2, 1),
(c) (1,1,1), (d) C3, -2,1).
9. Find the positions of the following planes relative to the sphere
x2 4-«/2+02 + 4x-3?/ + 6« + 5 = O:
(a) 4:X-\-2y + z + 2 = 0, (b)Sx-y-^z + 6 = 0.
10. Find the positions of the following lines relative to the sphere of
Ex. 9 : (a)2x-y + 2z + 7 = 0, Sx- y-z -10 = 0.
(by Bx + 8y + z -9=0, x-8y-{-z-\-n = 0.
11. Find the coordinates of the ends of that diameter of the sphere
x^ + y^ + z'^ — 6x — 6y-\-4iZ — QQ = 0, which lies on the line joining the
origin and the center.
XV, § 342] THE SPHERE 321
341. Equations of a Circle. In solid analytic geometry a
curve is represented by two simultaneous equations (§ 310),
that is, by the equations of any two surfaces intersecting in
the curve. Thus two linear equations represent together the
line of intersection of the two planes represented by the two
equations taken separately (§§322, 326).
A linear equation together with the equation of a sphere,
^ ^ x^ ■\- y"^ ■\- z^ ^ ax -\- by -\- cz -\- d = (),
represents the locus of all those points, and only those points,
which the plane and sphere have in common. Thus, if the
plane intersects the sphere, these simultaneous equations rep-
resent the circle in which the plane cuts the sphere; if the
plane is tangent to the sphere, the equations represent the
point of contact; if the plane does not intersect or touch
the sphere, the equations are not satisfied simultaneously by
any real point.
342. Sections Perpendicular to Axes. Projecting Cylinders.
In particular, the simultaneous equations
(4) z = Tc, ic2 + 2/2 + ;s2 ^ 7-2
represent, if A: < r, a circle about the axis Oz {i.e. a circle
whose center lies on Oz and whose plane is perpendicular to
Oz). If the value of z obtained from the linear equation be
substituted in the equation of the sphere, we obtain an equation
in X and y, viz. „ „ „ , «
which represents (since z is arbitrary) the circular cylinder,
about Oz as axis, which projects the circle (4) on the plane
Oxy. Interpreted in the plane Oxy, i.e. taken together with
2 = 0, this equation represents the projection of the circle (4)
on the plane Oxy.
Similarly if we eliminate x ov y ot z between the equations
322 SOLID ANALYTIC GEOMETRY [XV, § 342
(3) we obtain an equation in y and z, z and x, or x and ?/, rep-
resenting the cylinder that projects the circle (3) on the plane
Oyz, Ozx, or Oxy, respectively.
343. Tangent Plane. The tangent plane to a sphere at any
point Pi of the sphere is the plane through P^ at right angles
to the radius through Pj .
For a sphere whose center is at the origin,
^ + y"^ -\-z'^ = r^,
the equation of the tangent plane at P\{x-^, yi, Zi) is found by
observing that its distance from the origin is r and that the
direction cosines of its normal are those of OPi, viz. Xi/r,
yi/r,Zi/r. Hence the equation
(5) x^x + 2/i2/ + ^iZ = r\
If the equation of the sphere is given in the general form
A{x^ +y''+z'')+2 0x + 2Hy -\-2Iz ^ J=0,
we obtain by transforming to parallel axes through the center
the equation
the tangent plane at P^ix^, 2/1 ? %) ^^^^^ is
a^ia^ + 2/12/4- ^1^ = ^ + ^, + ^--
Transforming back to the original axes, we have :
(--S(-S)*('-!)("!)H-3(-i)
A^ A^ A^ a'
Multiplying out and rearranging, we find that the equation of
the tangent plane to the sphere
Aix" + y"" + z'') + 2 Qx + 2 Hy -\-2 Iz + J =
at the point Pi (fl?i, 2/1? ^\) is
(6) A{x^x-\-yiy+z^z)^-0(x,^x) +H{y,^y)+I{z^ + z)+ J= 0.
XV, § 344] THE SPHERE 323
344. Intersection of Line and Sphere. The intersections
of a sphere about the origin,
x^ -\- y^ -{- z"^ = r^,
with a line determined by two of its points Pi(xi,yi, Zi) and
PgC^aj Vi) ^2)) and given in the parameter form [(5), § 328]
x = Xi + k{x2-x{), y = yi-\-Jc(y2-yi), z=z^-\-'k{z2-z{),
are found by substituting these values of ic, ?/, z in the equation
of the sphere and solving the resulting quadratic equation in k :
[x, + J€(x, - x,)Y + [?A + k(y2 - yi)Y + [^1 + k(z, - z{)Y = r\
which takes the form
\_(x^ - a;i)2 -h {y, - y^y + (z^ - z^y^ k^ + 2 [x, {x^ - x,) -\- y^ {y^ - y,)
The line P1P2 will intersect the sphere in
two different points, be tangent to the
sphere, or not meet it at all, according as
the roots of this equation in k are real and
different, real and equal, or imaginary ; i.e.
according as
where d denotes the distance of the points Pi and Pg. Divid-
ing by d^, we can write this condition in the form
r' - \x,^ + 2/1^ + z,^ - U""-^' + 2/1^^' + ^i^-^'Y] I ^>
where by § 334 the quantity in square brackets is the square
of the distance 8 from the line P1P2 to the origin (Fig. 139).
Our condition means therefore that the line P1P2 meets the
sphere in two different points, touches it, or does not meet it
at all according as
which is obvious geometrically.
324 SOLID ANALYTIC GEOMETRY [XV, § 345
345. Tangent Cone. The condition for the line P^P^ to be
tangent to the sphere is (§ 344) :
W+ yi'+z,'-r')l(x,-x,y + (2/2 - ^i)^ + (^2 - ^i)'].
To give this expression a more symmetric form let us put, to
abbreviate,
X1X2 + 2/1^2 + 2;i2;2 = p, a^i' + 2/1' + ^i" = gu ^2' + 2/2' 4- ^2' = 92,
so that the condition is
(p-qiY. = (qi-r'){q,-2p + q2)y
i.e. p^ — 2 r^p = q^q^ — r'^q^ — r^^g 5
adding r* in both members, we have
i.e.
{x^X2 + 2/12/2 + z,z^ - r')' = (a?i' -f 2/1' + =2i' - r'){x^^ + 2/2' + ^2' - r").
Now keeping the sphere and the point Pj fixed, let Pg vary
subject only to this condition, i.e. to the
condition that P^P^ shall be tangent to
the sphere; the point Pg, which we shall
now call P{x^ y, z) is then any point of
the cone of vertex Pj tangent to the cone. ?i
Hence the equation of the cone of vertex Fia. 139
-f*i(^i f 2/1 J ^1) tangent to the sphere x^ + y^ -\- z'^ = r^ is
i^i' + 2/1' + ^1' - r'Xx' + 7f-\-z'-r') = {xix + 2/12/ + z,z - r^f.
If, in particular, the point Pi is taken on the sphere so that
^\ + yi + z-^ = r^, the equation of the tangent cone reduces to
the form ^,^ + y,y + ,,, = ^,
which represents the tangent plane at P^.
346. Inversion. A sphere of center and radius a being given,
we can find to every point P of space (excepting 0) one and only one
XV, § 346] THE SPHERE 325
point P' on OP (produced if necessary) such that
OP' OP' = aK
The points P, P' are said to be inverse to each other with respect to the
sphere (compare § 91).
Taking rectangular axes through 0, we find as the relations between
the coordinates of the two inverse points P(x, y, z) and P'{x\ y', z') if
we put OP = r = Va:2 + y^ + z^. OP' = r' = v'x'=^ + y'^ + z'^ :
x_y' _z' _r' _ rr' _ a^ .
X y z r f^ r'^ '
hence x' - ^"^ y'-—^ z' - ^ •
hence ^-^2 + ^2 + ^2' ^"^2 + ^2 + ^2' ^-:,2 + ^2 + ^2'
and similarly
y =
x'-^ -^ y'^ -}- z'^ x'-2 + y'-^ + z'-^ x''^ + y'^ + z'-^
These equations enable us to find to any surface whose equation is given
the equation of the inverse surface, by simply substituting for x, y, z
their values.
Thus it can be shown, that by inversion every sphere is transformed
into a sphere or a plane. The proof is similar to the corresponding propo-
sition in plane analytic geometry (§ 92) and is left as an exercise.
EXERCISES
1. Find the radius of the circle which is the intersection : (a) of the
plane y = Q with the sphere x^ + y^ -\- z^ — 6y = ; (6) of the plane
2x—Sy + z-2 = with the spherfe x^ + y'^ + z^ -6x + 2y - lb = 0.
2. A line perpendicular to the plane of a circle through its center is
called the axis of the circle. Find the circle : (a) which lies in the plane
z = 4:, has a radius 3 and Oz as axis ; (6) which lies in the plane 2/ = 5,
has a radius 2 and the line x — 3 = 0, — 4=0 as axis.
3. Find the circles of radius 3 on the sphere of radius 4 about the
origin whose common axis is equally inclined to the coordinate axes.
4. Does the line joining the points (2, — 1, — 6), (- 1, 2, 3) intersect
the sphere x^ + y'^ + z^ = 10? Find the points of intersection.
326 SOLID ANALYTIC GEOMETRY [XV, § 346
5. Find the planes tangent to the following spheres at the given
points : (a) x'^ + y^ -{- z'^ -Sy - 5z ~2 = 0, at (2, - 1, 3) ;
(6) a;2 + 2/2 4. 2.2 _|_ 2 X - 6 y + 2! -1 = 0, at (0, 1, - 3) ;
(c) S{x^-\-y^ + z^)-5x + 2y - z = 0, at the origin ;
(d) a;2 + y2 ^z'^^ax- by -cz = 0, at (a, 6, c).
6. Find the tangent cone : (a) from (4, 1, — 2) to x^ -\- y^ -\- z^ = Q ;
(&) from (2 a, 0, 0) to x^ + y^ + z^ = a^] (c) from (4, 4, 4) to x^ + if
+ «2 _ 16 ; (e^) from (1, - 5, 3) to x^ + y^-\-z'^ = 9.
7. Find the cone with vertex at the origin tangent to the sphere
(x-2ay-\-y^ + z^ = a^.
8. Show that, by inversion with respect to the sphere x^ -\- y'^ + ^2 _ ^52^
every plane (except one through the center) is transformed into a sphere
passing through the origin.
9. With respect to the sphere x"^ -}■ y^ + z^ = 25, find the surfaces in-
verse to (a) x = 6, (6) x-y = 0, (c) 4 (x^ + y^ -\- z^)-20 x-25 = 0.
10. Show that by inversion with respect to the sphere ^2 -]- y^ -}■ z^ = cfi
every line through the origin is transformed into itself.
11. With respect to the sphere x'^ -\-y'^ -\- z^ = a^, find the surface in-
verse to the plane tangent at the point Pi (xi , yi , Zi).
12. Show that all spheres with center at the center of inversion are
transformed into concentric spheres by inversion.
13. What is the curve inverse to the circle a;2 -f y2 _|_ ^2 _ 25, = 4,
with respect to the sphere a;2 + ^2 _|_ ^2 _ iq 9
347. Poles and Polars. Let P and P> be inverse points with
respect to a given sphere ; then the plane tt through P', at right angles to
OP ( being the center of the sphere) , is called the polar plane of the
point P, and P is called the pole of the plane tt, with respect to the
sphere.
With respect to a sphere of radius a, with center at the origin^
ic2 + 2/2 + 2^2 = a^,
the equation of the polar plane of any point P\ {x\ , y^ Z\) is readily
found by observing that its distance from the origin is a2/ri, and that the
XV, § 349] THE SPHERE • 327
direction cosines of its normal are equal to xi/n, yi/n, z\fr\<^ where
r^ = xi'^^- y-^ + Z'^ ; the equation is therefore
x\x + y\y + z\.z = a2.
If, in particular, the point Pi lies on the sphere, this equation, by § 343
(5), represents the tangent plane at Pi. Hence the polar plane of any
point of the sphere is the tangent plane at that point ; this also follows
from the definition of the polar plane.
348. With respect to the same sphere the polar planes of any two
points Pi(a;i , yi , zi) and P2(iC2 , 1/2 , Z2) are
xix + yiy + ziz = a2 and X2X + y2y + z^z = a^.
Now the condition for the polar plane of Pi to pass through P2 is
a^ia;2 + ym + Z1Z2 = <jfi ;
but this is also the condition for the polar plane of P2 to pass through Pi.
Hence the polar planes of all the points of any plane w (not passing
through the origin 0) pass through a common point, namely, the pole
of the plane ir ; and conversely, the poles of all the planes through a com-
mon point P lie in a plane, namely, the polar plane of P.
349. The polar plane of any point P of the line determined by two
given points Pi(xi , yi , zi) and P2(a;2 , t/2 , Z2) (always with respect to the
same sphere x^ -{■ y^ + z^ = a^) is
Ixi + k(X2 -xi)]x-{- [yi + k(^y2 - yi)]y + [zi+ k{z2 - zi)'[z = a\
This equation can be written in the form
k.
xix + yxy + ziz — a'^ + - — - {X2X + y2y + Z2Z — a^) = 0,
1 — fC
which for a variable k represents the planes of the pencil whose axis is the
intersection of the polar planes of Pi and P2. Hence the polar planes of
all the points of a line X pass through a common line ; and conversely,
the poles of all the planes of a pencil lie on a line.
Two lines related in this way are called conjugate lines (or conjugate
axes, reciprocal polars). Thus the line P1P2
x-x\ _ y —yx _ z - z\
X2 - Xi 2/2 — yx Z2 — Zx
328 SOLID ANALYTIC GEOMETRY [XV, § 349
and the line xix + yiy + ziz = a'^,
X2X + ViV + ZiZ = a^
are conjugate with respect to the sphere x^ + y2 ^ ^2 _ ^^2,
As the direction cosines of these lines are proportional to
X2 — X1, yi- y\, ^2 - zi
and
yi zi
2/2 ^2
\x\ y\\
I Xi yi \
Z\ X\
Z2 Xl
respectively, the two conjugate lines are at right angles (§ 331).
350. By the method used in the corresponding problem in the plane
(§ 95) it can be shown that the polar plane of any point P\{x\ , y\ , z\)
with respect to any sphere
^(^2 + 1/2 _|_ 2;2) + 2 G^X + 2 ^2/ + 2 /^ + ^"1=
is
A{xxx + yxy + zxz) + G{xx + x) + H{yx + y) + I{zx + 0) + jr = 0.
351. Power of a Point, if in the left-hand member of the equation
of the sphere
(X - Kf + (2/ - j)2 + (^ - kY - r2 =:
we substitute for x, ?/, 0, the coordinates xi , yi , ^1 of any point not on
the sphere, we obtain an expression (xi — uy + (yi — j)2+ (s^i — A;)2 — r2
different from zero which is called the power of the point Pi (xi ^ yi, zi^
with respect to the sphere.
As (xi — 7i)2 + (yi — j)2 + (zi - A:)2 is the square of the distance d be-
tween the point Pi and the center C of the sphere, we can write the
power of Pi briefly
<Z2 - r2 ;
the power of Pi is positive or negative according as Pi lies outside or
within the sphere. For a point Pi outside, the power is evidently the
square of the length of a tangent drawn from Pi to the sphere.
352. Radical Plane, Axis, Center. The locus of a point whose
powers with respect to the two spheres
a:2 + ?/2 + 22 + a^x + biy + ciz + cZi = 0,
x^+y^ + z^ + aix + biy + C2Z ^ d2 = Q
are equal is evidently the plane
(ai — a2)x + (5i — h2)y + (ci - C2)z + tZi — ^2 = 0,
which is called the radical plane of the two spheres. It always exists un-
less the two spheres are concentric.
XV, §353] THE SPHERE 329
It is easily proved that the three radical planes of any three spheres
(no two of which are concentric) are planes of the same pencil (§ 323) ;
and hence that the locus of the points of equal power with respect to
three spheres is a straight line. This line is called the radical axis of the
three spheres ; it exists unless the centers lie in a straight line.
The six radical planes of four spheres, taken in pairs, are in general
planes of a sheaf (§ 324) . Hence there is in general but one point of
equal power with respect to four spheres. This point, the radical center
of the four spheres, exists unless the f our'centers lie in a plane.
353. Family of Spheres. The equation
represents a family^ or pencil, of spheres^ provided k ^—1. If the two
spheres
x2 + 2/2 + z^ + «ix + hiy + ciz + (^i = 0,
X2 + ?/2 + 2r2 + a2.X + b^y + C2^ -h ^2 =
intersect, every sphere of the pencil passes through the common circle of
these two spheres. If ^• = — 1, the equation represents the radical plane
of the two spheres.
EXERCISES
1. Find the radius of the circle in which the polar plane of the point
(4, 3, — 1) with respect to x:^-\-y'^-\-z^ = 16 cuts the sphere.
2. Find the radius of the circle in which the polar plane of the point
(5, — 1, 2) with respect to x'^ -{■ y'^ + z"^ — 2x + ^y = ^ cuts the sphere.
3. Show that the plane 3ic + ?/— 4s = 19 is tangent to the sphere
x'^ + y^ + z'^ — 2x — ^y — Qz— \2,=0^ and find the point of contact.
4. If a point describes the plane 4 x — 5 ?/ — 3 a: = 16, find the coordi-
nates of that point about which the polar plane of the point turns with
respect to the sphere aj2 -f y2 ^ ^2 = 16.
5. If a point describes the plane 2a: + 3y + 5! = 4, find that point
about which the polar plane of the point turns with respect to the sphere
x2 + y2 _|_ 2-2 _ 8.
6. If a point describes the line ^ ~ = ^-i— = ^ ~ , find the equa-
o 5 — 2i
tions of that line about which, the polar plane of the point turns with
330 SOLID ANALYTIC GEOMETRY [XV, § 353
respect to the sphere x^ + y^ + z^ = 25. Show that the two lines are
perpendicular.
7. If a point describe the line 2x-Sy-\-iz = 2, x + y -{- z = S, find
the equations of that line about which the polar plane of the point turns
with respect to the sphere x"^ + y^ -{- z'^ = 16. Show that the two lines are
perpendicular.
8. Find the sphere through the origin that passes through the circle
of intersection of the spheres x^ -\- y'^+z"^ — 3 x -^ i y — 6 z — 8 = 0, x^-\-y'^
-{- z^ - 2 X + y ~ z - 10 = 0.
9. Show that the locus of a point whose powers with respect to two
given spheres have a constant ratio is a sphere except when the ratio is
unity.
10. Show that the radical plane of two spheres is perpendicular to the
line joining their centers.
11. Show that the radical plane of two spheres tangent internally or
externally is their common tangent plane.
12. Find the equations of the radical axis of the spheres x^ -\- y^+ z^
-Sx-2y -z-^ = 0, x'^+y^ + z^+5x-Sy-2z-S = 0, x^ + ^a
+ 02 _ 16 = 0.
13. Find the radical ^center of the spheres x^ -\-y'^ + z^ — 6 x -{- 2y
- ;2 + 6 = O; 5C2 + ?/2 + 02 _ 10 = 0, X2 + 1/2 + 02 + 2 x - 3 y + 5 2! - 6 = 0,
x^ + y^ -{- z^ - 2x + 4 y - 12 =0.
14. Show that the three radical planes of three spheres are planes of
the same pencil.
15. Two spheres are said to be orthogonal when their tangent planes
at every point of their circle of intersection are perpendicular. Show
that the two spheres x^ -}-y^-^ z^ + aix + biy + Ciz + t^i = 0, x2 -\- y"^ + z'^
+ a^x + hiy + C20 + 0^2 = are orthogonal when aia^ + 6160 + C1C2
= 2(di + (?2).
16. Write the equation of the cone tangent to the sphere x^ + y'^ ■\-
-5.2 — 1.2 ^itii vertex (0, 0, z\). Divide this equation by zi^ and let the
vertex recede indefinitely, i.e. let 01 increase indefinitely. The equation
a;2 4- 2/2 = |.2^ thus obtained, represents the cylinder with axis along the
axis Oz and tangent to the sphere x^ 4- y'^ -\- z^ = r'^.
XV, § 353] THE SPHERE 331
17. In the equation of the tangent cone (§ 345) write for the
coordinates of the vertex xi = nh , l/i = nnii , zi = rini ; divide the equa-
tion by n^ and let n increase indefinitely, i.e. let the vertex of the cone
recede indefinitely. The tangent cone thus becomes a tangent cylinder
with axis passing through the center of the sphere and having the direc-
tion cosines h, mi, ni. Show that this tangent cylinder is
(hx + miy + nizy^ - {x^ + y^ + z^ - r^) = 0.
18. From the result of Ex. 17, find the cylinder with axis equally
inclined to the coordinate axes which is tangent to the sphere x^ -\- y^
-\-z^ = r2.
19. From the result of Ex. 17, find the cylinders with axes along the
coordinate axes which are tangent to the sphere x'^ -{-y"^ + z^ = r^.
20. Find the cylinder with axis through the origin which is tangent to
the sphere x^ -{-y'^ + z'^ — 4:X + 6y — S = 0.
21. Find the family of spheres inscribed in the cylinder
{Ix + my-\- nzy - (x?- -\-y^-\-z'^- r^) = 0.
22. Find the cylinder with axis having direction cosines Z, m, n which
is tangent to the sphere (x — h)'^ +(y — j)'^ -\-{z — k)^ = r^.
23. Show that as the point P recedes indefinitely from the origin along
a line through the origin of direction cosines Z, m, n, the polar plane of P
with respect to the sphere x^ -f y'^ + z^ = a^ becomes ultimately Ix + my
+ nz = 0.
CHAPTER XVI
QUADRIC SURFACES
354. The Ellipsoid. The surface represented by the
equation
is called an ellipsoid. Its shape is best investigated by tak-
ing cross-sections at right angles to the axes of coordinates.
Thus the coordinate plane Oyz whose equation is ic = in-
tersects the ellipsoid in the ellipse
Any other plane perpendicular to the axis Ox (Fig. 140), at
y
Fig. 140
the distance h ^ a from the plane Oyz intersects the ellipsoid
in an ellipse whose equation is
7i'
^2 + ^2--^ .2'
I.e.
f
K'-S) -(•--:)
= 1.
332
XVI, § 355] QUADRIC SURFACES 333
Strictly speaking this is the equation of the cylinder that pro-
jects the cross-section on the plane Oyz. But it can also be
interpreted as the equation of the cross-section itself, referred
to the point Qi, 0, 0) as origin and axes in the cross-section
parallel to Oy and Oz.
Notice that as h < a, Jv^/a^, and hence also 1 — h^/a"^, is a posi-
tive proper fraction. The semi-axes 6Vl — h^/a^, c VI — h'^/a'^
of the cross-section are therefore less than b and c, respec-
tively. As h increases from to a, these semi-axes gradually
diminish from b, c to 0.
355. Cross-Sections. Cross-sections on the opposite side
of the plane Oyz give the same results ; the ellipsoid is evi-
dently symmetric with respect to the plane Oyz.
By the same method we find that cross-sections perpendicu-
lar to the axes Oy and Oz give ellipses with semi-axes dimin-
ishing as we recede from the origin. The surface is evidently
symmetric to each of the coordinate planes. It follows that
the origin is a center, i.e. every chord through that point is
bisected at that point. In other words, if (x, y, z) is a point
of the surface, so is {—x, —y, —z). Indeed, it is clear from
the equation that if {x, y, z) lies on the ellipsoid, so do the
seven other points {x, y, -2), {x, -y, z), {-x, y, z), {x, -y, -z),
(—x,y, —z), (—X, —y, z), {—x,—y,—z). A chord through
the center is called a diameter.
It follows that it suffices to study the shape of the portion of
the surface contained in one octant, say that contained in the tri-
hedral formed by the positive axes Ox, Oy, Oz ; the remaining
portions are then obtained by reflection in the coordinate planes.
The ellipsoid is a dosed surface; it does not extend to in-
finity ; indeed it is completely contained within the parallel-
epiped with center at the origin and edges 2 a, 2 &, 2 c, parallel
to Ox, Oy, Oz, respectively.
334
SOLID ANALYTIC GEOMETRY [XVI, § 356
356. Special Cases. In general, the semi-axes a, h, c of the
ellipsoid, i.e. the intercepts made by it on the axes of coordi-
nates, are different. But it may happen that two of them, or
even all three, are equal.
In the latter case, i.e. if a = h = c, the ellipsoid evidently
reduces to a sphere.
If two of the axes are equal, e.g. if & = c, the surface
a" ¥ b^
Fig. 141
is called an ellipsoid of revolution because it can be generated
by revolving the ellipse
y\
about the axis Ox (Fig. 141).
Any cross-section at right angles
to Ox, the axis of revolution, is a ^
circle, while the cross-sections at
right angles to Oy and Oz are
ellipses. The circular cross-section in the plane Oyz is called
the equator; the intersections of the surface with the axis of
revolution are the poles.
If a > 6 (a being the intercept on the axis of revolution),
the ellipsoid of revolution is called prolate; if a < b, it is
called oblate. In astronomy the ellipsoid of revolution is
often called spheroid, the surfaces of the planets which are
approximately ellipsoids of revolution being nearly spherical.
Thus for the surface of the earth the major semi-axis, i.e. the
radius of the equator, is 3962.8 miles while the minor semi-
axis, i.e. the distance from the center to the north or south
pole, is 3949.6 miles.
XVI, § 357]
QUADRIC SURFACES
335
367. Surfaces of Revolution. A surface that can be gen-
erated by the revolution of a plane curve about a line in the
plane of the curve is called a surface of revolution. Any such
surface is fully determined by the generating curve and the
position of the axis of revolution with respect to the curve.
Let us take the axis of revolution as axis Ox, and let the
equation of the generating curve be
As this curve revolves about Ox, any
point P of the curve (Fig. 142) de-
scribes a circle about Ox as axis,
with a radius equal to the ordinate
f{x) of the generating curve. For
any position of P we have therefore
and this is the equation of the surface of revolution.
Thus if the ellipse
a^ h-'
revolves about the axis Ox, we find since y = ± (b/a) Va'
for the ellipsoid of revolution so generated the equation
Fig. 142
a;2
a
x^),
which agrees with that of § 356.
Any section of a surface of revolution at right angles to the
axis of revolution is of course a circle ; these sections are called
parallel circles, or simply parallels (as on the earth's surface).
Any section of a surface of revolution by a plane passing
through the axis of revolution is called a meridian section ;
it consists of the generating curve and its reflection in the axis
of revolution.
336 SOLID ANALYTIC GEOMETRY [XVI, § 357
EXERCISES
— 1. An ellipsoid has six /oci, viz. the foci of the three ellipses in which
the ellipsoid is intersected by its planes of symmetry. Determine the
coordinates of these foci : (a) for an ellipsoid with semi-axes 1, 2, 3 ;
(6) for the earth (see §356) ; (c) for an ellipsoid of semi-axes 10, 8, 1 ;
(d) for an ellipsoid of semi-axes 1, 1, 5.
2. Show that the intersection of an ellipsoid with any plane actually
cutting the ellipsoid is an ellipse by proving that the projection of this
curve of intersection on each coordinate plane is an eUipse.
3. Assuming a > & > c in the equation of § 354 find the planes through
Oy that mtersect the ellipsoid in circles.
'- " 4. Find the equation of the paraboloid of revolution generated by the
revolution of the parabola y'^ = 4: ax about Ox.
6. Find the equation of a torus, or anchor-ring, i.e. the surface
generated by the revolution of a circle of radius a about a line in its plane
at the distance b> a from its center.
6. Find the equation of the surface generated by the revolution of a
. circle of radius a about a line in its plane at the distance & < a from its
center. Is the appearance of this surface noticeably different from the
surface of Ex. 5 ?
7. Show what happens to the surface of Ex. 6 when 6 = 0; when & = a.
8. Find the equation of the surface generated by the revolution of the
parabola y^ = 4tax about: (a) the tangent at the vertex; (&) the latus
rectum.
"* 9. Find the equation of the surface generated by the revolution of the
hyperbola xy = a^ about an asymptote.
10. Find the cone generated by the revolution of the line y = mx -{- b
about: (a) Ox, (6) Oy.
11. How are the following surfaces of revolution generated ?
(a) y^+z^=x^. (&) 2x^+2y^-^z=0. (c) x^-\-y^-z'^-2x+i = 0.
12. Find the equation of the surface generated by the revolution of
the ellipse x^ + 4 ?/2 — 4 x = : (a) about the major axis ; (b) about the
minor axis ; (c) about the tangent at the origin.
XVI, § 359]
QUADRIC SURFACES
337
358. Hyperboloid of One Sheet. The surface represented
by the equation
= 1
a" y^ &
is called a hyperboloid of one sheet (Fig. 143). The intercepts
Fig. 143
on the axes Ox, Oy are ±a, ± 6 ; the axis Oz does not intersect
the surface.
359. Cross-Sections. The plane Oxy intersects the surface
in the ellipse
cross-sections perpendicular to Oz give ellipses with ever-in-
creasing semi-axes.
The planes Oyz and Ozx intersect the surface in the hyperbolas
Any plane perpendicular to Ox, at the distance h from the
origin, intersects the hyperboloid in a hyperbola, viz.
f ■ z'
338 SOLID ANALYTIC GEOMETRY [XYI, § 359
As long as /i < a this hyperbola has its transverse axis parallel
to Oy while for h > a the transverse axis is parallel to Oz ; for
h = a the equation reduces to y'^/h'^ — z^/c^ = and represents
two straight lines, viz. the parallels through (a, 0, 0) to the
asymptotes of the hyperbola y^/b"^ — z^/x^ = 1 which is the
intersection of the surface with the plane Oyz.
Similar considerations apply to the cross-sections perpen-
dicular to Oy.
The hyperboloid has the same properties of symmetry as the
ellipsoid (§ 355) ; the origin is a center, and it suffices to inves-
tigate the shape of the surface in one octant.
360. Hyperboloid of Revolution of one Sheet. If in the
hyperboloid of one sheet we have a = b, the cross-sections per-
pendicular to the axis Oz are all circles so that the surface can
be generated by the revolution of the hyperbola
about Oz. Such a surface is called a hyperboloid of revolution
of one sheet.
361. Other Forms. The equations
^-^' + - = 1 -^ + ^' + ?! = l
a2 b^ d" ' a^ b"" c^
also represent hyperboloids of one sheet which can be investi-
gated as in §§ 358-360. In the former of these the axis Oy, in
the latter the axis Ox, does not meet the surface.
Every hyperboloid of one sheet extends to infinity.
362. Hyperboloid of Two Sheets. The surface represented
by the equation
a"- ¥ c2~
is called a hyperboloid of two sheets (Fig. 144).
XVI, § 365]
QUADRIC SURFACES
339
The intercepts on Ox are ± a ; the axes Oy, Oz do not meet
the surface.
363. Cross-Sections. The cross-sections at right angles to
Ox, at the distance h from the origin are
2/' 2;2 _ .
(-S)
C2fl
these are imaginary as long as 7i < a;
for h>a they are ellipses with ever-
increasing semi-axes as we recede from
the origin.
The cross-sections at right angles to Oy
and Oz are hyperbolas.
The hyperboloid of two sheets, like that of one sheet and
like the ellipsoid, has three mutually rectangular planes of
symmetry whose intersection is therefore a center.
The surfaces
Fig. 1M
_^ _,]r ^—i ^^^yiA-^—i
are hyperboloids of two sheets, the former being met by Oy,
the latter by Oz, in real points.
The hyperboloid of two sheets extends to infinity.
364. Hyperboloid of Revolution of Two Sheets. If & = c
in the equation of § 362, the cross-sections at right angles to Ox
are circles and the surface becomes a hyperboloid of revolution
of two sheets. .
365. Imaginary Ellipsoid. The equation
_x^ _y^ _z^ _A
is not satisfied by any point with real coordinates. It is some-
times said to represent an imaginary ellipsoid.
340 SOLID ANALYTIC GEOMETRY [XVI, § 366
366. The Paraboloids. The surfaces
a;2 7/2_
a- 0^ a^ W
which are called the elliptic paraboloid (Fig. 145) and hyper-
bolic paraboloid (Fig. 146), respectively, have each only two
planes of symmetry, viz the planes Oyz and Ozx. We here
assume that c^O. The cross-sections at right angles to the
Fig. 145
Fig. 146
axis Oz are evidently ellipses in the case of the elliptic parab-
oloid, and hyperbolas in the case of the hyperbolic paraboloid.
The plane Oxy itself has only the origin in common with the
elliptic paraboloid ; it intersects the hyperbolic paraboloid in
the two lines x'^/a'^ — y'^/b^ = 0, i.e. y = ± bx/a.
The intersections of the elliptic, paraboloid (Fig. 145) with
the planes Oyz and Ozx are parabolas with Oz as axis and as
vertex, opening in the sense of positive 2 if c is positive, in the
sense of negative 2; if c is negative. Planes parallel to these
coordinate planes intersect the elliptic paraboloid in parabolas
with axes parallel to Oz, but with vertices not on the axes Ox,
Oy, respectively.
For the hyperbolifc paraboloid (Fig. 146), which is saddle-
shaped at the origin, the intersections with the planes Oyz and
XVI, § 369]
QUADRIC SURFACES
341
Ozx are also parabolas with Oz as axis ; if c is positive the
parabola in the plane Oyz opens in the sense of negative z, that
in the plane Ozx opens in the sense of positive z. Similarly
for the parallel sections.
367. Paraboloid of Revolution. If in the equation of the
elliptic paraboloid we have a=b, it reduces to the form
x^-\-y^ = 2pz.
This represents a surface of revolution, called the paraboloid of
revolution. This surface can be regarded as generated by the
revolution of the parabola y'^ = 2pz about the axis Oz.
368. Elliptic Cone. The surface represented by the equation
x^ y^
=
is an elliptic cone, with the origin as vertex and the axis Oz as
axis (Fig. 147).
The plane Oxy has only the origin in
common with the surface. Every parallel
plane z = k, whether Jc be positive or negative,
intersects the surface in an ellipse, with
semi-axes increasing proportionally to k.
The plane Oyz, as well as the plane Ozx,
intersects the surface in two straight lines
through the origin. Every plane parallel to
Oyx or to Ozx intersects the surface in a
hyperbola. Fig. 147
369. Circular Cone. If in the equation of the elliptic cone
we have a = b, the cross-sections at right angles to the axis Oz
become circles. The cone is then an ordinary circular cone, or
342 SOLID ANALYTIC GEOMETRY [XVI, § 369
cone of revolution, which can be generated by the revolution
of the line y = (^a/c)z about the axis Oz. Putting a/c = m we
can write the equation of a cone of revolution about Oz, with
vertex at 0, in the form
370. Quadric Surfaces. The ellipsoid, the two hyper-
boloids, the two paraboloids, and the elliptic cone are called
quadric surfaces because their cartesian equations are all of
the second degree.
Let us now try to determine, conversely, all the various loci
that can be represented by the general equation of the second
degree
Ax^ + By^ + (7^2 + 2 Dyz + 2 Ezx + 2 Fxy
+ 2 6x-{-2Hy + 2Iz-{-J=0.
In studying the equation of the second degree in x and y
(§ 249) it was shown that the term in xy can always be
removed by turning the axes about the origin through a cer-
tain angle. Similarly, it can be shown in the case of three
variables that by a properly selected rotation of the coordinate
trihedral about the origin the terms in yx, zx, xy can in general
all be removed so that the equation reduces to the form
(1) Ax^ + Bi/^ + Cz^ +2Gx + 2Hy+2Iz-{-J-=0.
• This transformation being somewhat long will not be given
here. We shall proceed to classify the surfaces represented
by equations of the form (1).
371. Classification. The equation (1) can be further sim-
plified by completing the squares. Three cases may be distin-
guished according as the coefficients A, B, C are all three differ-
ent from zero, one only is zero, or two are zero.
XVI, §371] QUADRIC SURFACES 343
Case (a) : ^ ^ 0, jB ^ 0, C ^0. Completing the squares in
Xf y, z we find
Referred to parallel axes through the point (— G/A, — H/B,
— I/C) this equation becomes
(2) Ax''-\-Bf-]-Cz^ = J,.
Case (6) : A=^0,B^O, (7=0. Completing the squares in x
aud y we find
If /^ we can transform to parallel axes through the point
(—G/Aj — H/B, J2/2 1) so that the equation becomes
(3) Ax" + By^+2Iz = 0.
If, however, 7=0, we obtain by transforming to the point
(-G/A, -II/B,0)
(3') Aa^-\-By'=J,.
Case (c) : A^O, B = Of C = 0. Completing the square in
x we have
A(x-{-^\2Hy + 2Iz = ^-J=J,.
If H and I are not both zero we can transform to parallel
axes through the point (— G/A, J^/2 H, 0) or through (— G/A,
0, J3/2 /) and find
(4) Ax'-\-2Hy + 2Iz = 0.
If 7r= and /= we transform to the point (— G/A, 0, 0)
so that we find
(4') Ax'^J,,
344 SOLID ANALYTIC GEOMETRY [XVI, § 372
372. Squared Terms all Present. Case (a). We proceed to
discuss the loci represented by (2). If J^ 4^ 0, we can divide
(2) by J^ and obtain :
(a) if ^/t/i, 5/t7i, (7/Ji are positive, an ellipsoid (§ 354) ;
(fi) if two of these coefficients are positive while the third
is negative, a hyperboloid of one sheet (§ 358) ;
(y) if one coefficient is positive while two are negative, a
hyperboloid of tivo sheets (§ 362);
(8) if all three coefficients are negative, the equation is not
satisfied by any real point (§ 365) ;
If Ji = the equation (2) represents an elliptic cone (§ 368)
unless A, B, C all have the same sign, in which case the origin
is the only point represented.
373. Case (b). The equation (3) of §371 evidently fur-
nishes the two paraboloids (§ 366) ; the paraboloid is elliptic if
A and B have the same sign ; it is hyperbolic if A and B are of
opposite sign.
The equation (3') since it does not contains and hence leaves
z arbitrary represents the cylinder , with generators parallel to Oz,
passing through the conic Ax^ -f- By"^ = ./g. As A and B are
assumed different from zero, this conic is an ellipse if A/J2 and
and B/J2 are both positive, a hyperbola if A/J^ and B/Jc, are of
opposite sign, and it is imaginary if A/J2 and B/J<^ are both
negative. This assumes Jg ^ 0. If J^ = 0, the conic degen-
erates into two straight lines, real or imaginary ; the cylinder
degenerates into two planes if the lines are real.
374. Case (c). There remain equations (4) and (4'). To sim-
plify (4) we may turn the coordinate trihedral about Ox through
an angle whose tangent is — H/I-, this is done by putting
Bf + Hz' - Hy' + Iz'
x = x \ y= -^ z — — ^ ;
^H^+P v'H^ + P
XVI, §374] QUADRIC SURFACES 345
our equation then becomes
It evidently represents a parabolic cylinder, with generators
parallel to Oy.
Finally, the equation (4') is readily seen to represent two
planes perpendicular to Ox, real or imaginary, unless t/3 =
in which case it represents the plane Oyz.
EXERCISES
1. Name and locate the following surfaces :
{a) x^-{-2y^ + Sz^ = 4. (b) x^ + y^ - 5z - 6 = 0.
(c) x'^ - y^ -\- z^ = 4. (d) x^-y^ + z^ + Sz + 6 = 0.
(e) 2?/2 -4^2 _ 5=3 0. (/) 2x2 + 2/2 + 3^2 + 5 = 0.
(g) 6^2 + 2 x2 = 10. (h) z^-9 = 0.
(0 x2 - y + 1 = 0. (j) x^-'y^-z^ + Qz = 9.
(k) x2 + 3 ?/2 + 2;2 + 4 + 4 = 0. (I) z'^ + ?/;- 9 = 0.
2. The cone
x2/a-^ + 2/2/6-2 _ 2-2/02 =
is called the asymptotic cone of the hyperboloid of one sheet
x2/a2 + yl/yZ _ 22/c2 = 1.
Show that as z increases the two surfaces approach each other, i.e. they
bear a relation similar to a hyperbola and its asymptotes.
3. What is the asymptotic cone of the hyperboloid of two sheets ?
4. Show that the intersection of a hyperboloid of two sheets with any
plane actually cutting the surface is an ellipse, parabola, or hyperbola.
Determine the position of the plane for each conic.
5. Show that in general nine points determine a quadric surface and
that the equation may be written as a determinant of the tenth order
equated to zero.
6. Show that the surface inverse to the cylinder x'^ ■\- y"^ = a"^^ with
respect to the sphere ^2 + ^/^ + ^2 — ^2^ jg ^\^q torus generated by the rev-
olution of the circle {y — a/2y^ -\- z"^ = a?- about the axis Ox.
7. Determine the nature of the surface xyz = a^ by means of cross-
sections.
346 SOLID ANALYTIC GEOMETRY [XVI, § 375
375. Tangent Plane to the Ellipsoid. The plane tangent
to the ellipsoid
a" b^ c^
can be found as follows (compare §§ 344, 345). The equa-
tions of the line joining any two given points (x^, y^, z^) and
{^2,y2,^2) are
x = x^ + 'k{x^-x;)y y = yi-\-k(y^-yi), z = Zi-\-k(z2-Zi).
This line will be tangent to the ellipsoid if the quadratic
in k
o? ¥ (?
has equal roots. Writing this quadratic in the form
1_ a^ b^ c2 J
oFxiix^-x,) yifa-yi) I gife-gQ "];, I W I yi" .^i' i\=()
"^ [ a2 -^ 52 ^ c2 J ^\a''^b'^c\ J '
we find the condition
[(
" 52 ^2
_r (x^ - x,y O/2 - yiY _, fe - ZiY lf^i^ j_ ^ 4. 5l _ 1 V
\_ a" 6^ "^ c2 JVa2 "^ 62 ^ c2 J
If now we keep the point {Xi , 2/1 , 2!i) fixed, but let the point
(X2, yz, Z2) vary subject to this condition, it will describe the
cone, with vertex (x^ , 2/1 , ^i), tangent to the ellipsoid ; to indi-
cate this we shall drop the subscripts of X2, 2/2, Z2. If, in
particular, the point (xi , 2/1 , z^) be chosen on the ellipsoid, we
have
XVI, §377] QUADRIC SURFACES 347
and the cone becomes the tangent plane. The equation of the
tangent plane to the ellipsoid at the point {x^ , y^ , z^ is, therefore :
a" we'
376. Tangent Planes to Hyperboloids. In the same way
it can be shown that the tangent planes to the hyperboloids
a"" h^ c2~ ' a2 y- c2~
at (aji , 2/i , 2i) are
a2 62 f.1 ' ^2 ^2 ^2
By an equally elementary, but somewhat longer, calculation
it can be shown that the tangent plane to the quadric surface
Ax^ -\- By^ -\- Cz^ -{-2 Dyz -{-2 Ezx-^2 Fxy
-\-2Gx-\-2Hy + 2Iz^J=0
at (xi , 2/i , Zi) is :
AxiX + Byiy + Cz^z + D {y^z + z^y) + E (z^x + x^z) -\- F{x,y + ^/lO;)
-{.G(x,-{-x)-^ H{y, + y)-\- I(z, + z) + J= 0.
In particular, the tangent planes to the paraboloids
t + t^2cz, ''--t = 2cz
a" b^ ' a2 52
are
^2 ^ 52 ^ ' ^ ^' a2 62 V 1 -r ;
377. Ruled Surfaces. A surface that can be generated by
the motion of a straight line is called a ruled surface; the line
is called the generator.
The plane is a ruled surface. Among the quadric surfaces
not only the cylinders and cones but also the hyperboloid of
one sheet and the hyperbolic paraboloid are ruled surfaces.
348
SOLID ANALYTIC GEOMETRY [XVI, § 378
378. Rulings on a Hyperboloid of One Sheet. To show
this for the hyperboloid
a^ h"- & '
we write the equation in the form
62 c2
x^
and factor both members :
»+'
)e-9-^3('-3-
It is then apparent that any point whose coordinates satisfy
the two equations
^ + 5==;fcfl+^\ ^_^-l
1 -
/cV a
where A; is an arbitrary parameter, lies
on the hyperboloid. These two equa-
tions represent for every value of A: (^ 0)
a straight line. The hyperboloid of one
sheet contains therefore the family of
lines represented by the last two equa-
tions with variable A;.
In exactly the same way it is shown that the same hyper-
boloid also contains the family of lines
c V aj c k\ a J
Thus every hyperboloid of one sheet contains two sets of recti-
linear generators (Fig. 148).
Fig. 148
XVI, § 379]
QUADRIC SURFACES
349
379. Rulings on a Hyperbolic Paraboloid. The hyperbolic
paraboloid (Fig. 149)
also contains tivo sets of recti-
linear generators, namely,
^ + l = k.2cz, 2-1 = 1
a b a k
and
a b a b 7c'
Fig. 149
EXERCISES
1. Derive the equation of the tangent plane to :
(a) the elliptic paraboloid ; (b) the hyperbolic paraboloid ;
(c) the elliptic cone.
2. The line perpendicular to a tangent plane at a point of contact is
called the normal line. Write the equations of the tangent planes and
normal lines to the following quadric surfaces at the points indicated :
(a) xy9 + yyi - 02/16 = 1, at (3, - 1, 2) ;
(6) x2 + 2 2/2 + 02 = 10, at (2,1, -2);
(c) x2 + 2 ?/2 - 2 ^2 = 0, at (4, 1, 3) ; (d) x^-Sy^-z = 0, at the origin.
3. Show that the cylinder whose axis has the direction cosines I, m, n
and which is tangent to the ellipsoid od^a^ + y^/b^ -f z^/c^ = 1, is
W b'^ cy W b^ d'JW b'^ c^ I
4. Show that the plane Ix + my -{- nz = y/'V-a^ + ?n2&2 + ^12^2 jg tangent
to the ellipsoid x'^la'^ + 2/2/52 + ^ij^fi _ 1.
5. Show that the locus of the intersection of three mutually perpen-
dicular tangent planes to the ellipsoid x^ja'^ + ?/2/62 _^ ^2/^2 = 1, is the
sphere (called director sphere) x^ + y^ +z^ = a^ + 62 _|. ^2.
350 SOLID ANALYTIC GEOMETRY [XVI, § 379
6. Show that the elliptic cone is a ruled surface.
7. Show that any two linear equations which contain a parameter
represent the generating line of a ruled surface. What surfaces are gen-
erated by the following lines ?
(a) x-y + kz = 0,x + y-z/k = {i\ (6) 3 a; - 4 y ^ A:, (3 a;+4 y)k = \ ;
(c) X — y + 3 A-^ = 3 A;, k{x -\-y)— z = 3.
8. Show that every generating line of the hyperbolic paraboloid
or^/cfi — y'^b^ = 2 cz is parallel to one of the planes x^/a^ — y^/b^ = 0.
380. Surfaces in General. When it is required to deter-
mine the shape of a surface from its cartesian equation
the most effective methods, apart from the calculus, are the
transformation of coordinates and the taking of cross-sections,
generally (though not necessarily always) at right angles to
the axes of coordinates. Both these methods have been ap-
plied repeatedly to the quadric surfaces in the preceding
articles.
381. Cross-Sections. The method of cross-sections is ex-
tensively used in the applications. The railroad engineer de-
termines thus the shape of a railroad dam ; the naval architect
uses it in laying out his ship ; even the biologist uses it in con-
structing enlarged models of small organs of plants or animals.
382. Parallel Planes. When the given equation contains
only one of the variables x, y, z, it represents of course a set of
parallel planes (real or imaginary), at right angles to one of
the* axes. Thus any equation of the form
F{x)=0
represents planes at right angles to Ox, of which as many are
real as the equation has real roots.
XVI, § 386] QUADRIC SURFACES 351
383. Cylinders. When the given equation contains only two
variables it represents a cylinder at right angles to one of the
coordinate planes. Thus any equation of the form
F{x,y)=0
represents a cylinder passing through the curve F{x, y) = in
the plane Oxy, with generators parallel to Oz. If, in particular,
F(x, y) is homogeneous in x and y, i.e. if all terms are of the
same degree, the cylinder breaks up into planes.
384. Cones. When the given equation F(x, y, 2)=0 is
homogeneous in x, y, and z, i.e. if all terms are of the same
degree, the equation represents a general cone, with vertex at
the origin. For in this case, if {x, y, z) is a point of the sur-
face, so is the point (lex, ky, kz), where k is any constant; in
other words, if P is a point of the surface, then every point of
the line OP belongs to the surface ; the surface can therefore
be generated by the motion of a line passing through the origin.
385. Functions of Two Variables. Just as plane curves are
used to represent functions of a single variable, so surfaces can
be used to represent functions of two variables. Thus to obtain
an intuitive picture of a given function f{x, y) we may con-
struct a model of the surface
such as the relief map of a mountainous country. The ordi-
nate z of the surface represents the function.
386. Contour Lines. To obtain some idea of such a surface
by means of a plane drawing the method of contour lines or
level lines can be used. This is done, e.g., in topographical
maps. The method consists in taking horizontal cross-sections
at equal intervals and projecting these cross-sections on the hori-
zontal plane. Where the level lines crowd together the surface
is steep ; where they are relatively far apart the surface is flat.
352
SOLID ANALYTIC GEOMETRY [XVI, § 386
EXERCISES
1. What surfaces are represented by the following equations ?
(a) Ax-{-By+C = 0.
(c) y^-\-z^ = a^.
(e) zx = a^.
(g) x^-Sx^-x+S = 0.
(0 y = x^ - X - e.
(k) x^ + 2 ?/2 = 0.
(m) x'^-y^ = z^
(0) (x-l){y-2)(z-S) = 0.
(b) xcos^ -\- ysinp =p.
(d) z^-x^ = a^
(/) z^ = 4ay.^
(h) xyz = 0,
U) yz^-9y = 0.
(I) a;2 = yz.
(n) y2 + 2z'^-\-4zx = 0.
(p) a;3 -f y3 — 3 xyz = 0.
2. Determine the nature of the following surfaces by sketching the
contour lines :
(a) z=:x-{-y. (b) z = xy. (c)z = y/x. (d) z =x^ -^y^.
(e) z=x^-y^-\-4. (f)z = x^. (g) z=x'^-\-y^-4:X. (h)z = xy-x.
{i) z = 2\ (j) y=z'^-ix. {k)y = Sz^ + x^. {l)z=nx+y'\
3. The Cassinian ovals (§ 270) are contour lines of what surface ?
4. What can be said about the nature of the contour lines of a sur-
face z =f{x) ? Discuss in particular : (a) z = x^ — 9 ; (b) z = x^ — 8 ;
(c) y = z^ + 2z.
387. Rotation of Coordinate Trihedral. To transform the
equation of a surface from one coordinate trihedral Oxyz to another
Ox'y'z', with the same origin O, we
must find expressions for the old co-
ordinates X, y, z of any point P in terms
of the new coordinates x', y', z'. We
here confine ourselves to the case when
each trihedral is trirectangular ; this is
the case of orthogonal transformation,
or orthogonal substitution.
Let li, wi, wi, be the direction cosines
of the new axis Ox' with respect to the
old axes Ox, Oy, Oz (Fig. 150) ; similarly
h, m2, W2 those of Oy', and Z3, ma, W3 those of Oz'.
the scheme
Fig. 150
This is indicated by
XVI, §389] QUADRIC SURFACES 353
x'
y'
0'
h
h
h'
Wli
m2
mz
ni
W2
tiz
which shows at the same time that then the direction cosines of the old
axis Ox with respect to the new axes Ox', Oy' ^ Oz' are Zi, ^2, h, etc.
388. The nine direction cosines h, h, ••• n^ are sufficient to determine
the position of the new trihedral Ox'y'z' with respect to the old. But
these nine quantities cannot be selected arbitrarily ; they are connected by
six independent relations which can be written in either of the equivalent
forms
h^ + wii2 + n{^ = 1, ^2^3 + m^mz + ruiiz = 0,
(1) ^2^ + ^2'-^ + «2^ = 1, hh -h ms^ni + n^ni — 0,
h^ + rriz^ + W32 = 1, hh + Wim2 + W1W2 = 0,
or
h^ + h^ + ?3^ = 1, wini + W2W2 + wisws = 0,
(1') wii2 + W22 + W32 = 1, mh + n2h + nsh = 0,
Wl^ + W2^ + W32 = 1 , ZlWi + l2'm2 + ?3W»3 = 0.
The meaning of these equations follows from §§ 297 and 300. Thus
the first of the equations (1) expresses the fact that h, mi, ui are the
direction cosines of a line, viz. Ox' ; the last of the equations (1') ex-
presses the perpendicularity of the axes Ox and Oy ; and so on.
389. If X, y, z are the old, x', y', z' the new coordinates of one and
the same point, we find by observing that the projection on Ox of the
radius vector of P is equal to the sum of the projections on Ox of its
components x', y', z' (§ 294), and similarly for the projections on Oy
and Oz :
X = hx' + hy' + hz',
(2) y = mix' + mzy' + W30',
z = mx' + n^y' + n^z'.
Indeed, these relations can be directly read off from the scheme of
direction cosines in § 387.
Likewise, projecting on Ox', Oy', Oz', we find
x' = hx + miy + niz,
(2') yi = I2X + TO22/ -I- n2Z,
z' = Izx + m^y + mz.
2a
354 SOLID ANALYTIC GEOMETRY [XVI, § 389
As the equations (2), by means of which we can transform the equation
of any surface from one rectangular system of coordinates to any other
with the same origin, give x, y, z as linear functions of x',y', z', it follows
that such a transformation cannot change the degree of the equation of
the -surface.
390. Th-fe equation (2') must of course result also by solving the equa-
tions (2) for x', y', z', and vice versa. Putting
h h h
nil Wi2 wi3 = D,
ni Ui nz
solving (2) for x', y\ z', and comparing the coefficients of x, y, z with
those in (2') we find the following relations :
I)h = m^nz — W3W2, Bmi = n^h — Ush, Dni = hm^ — hm^i etc.
Squaring and adding the first three equations (compare Ex. 3, p. 45)
and applying the relations (1) we find : D^ = \.
By § 321, D can be interpreted as six times the volume of the tetrahe-
dron whose vertices are the origin and the points x', y', z' in Fig. 150, i.e.
the intersections of the new axes with the unit sphere about the origin.
The determinant gives this volume with the sign + or — according as the
trihedral Ox'y'z' is superposable or not (in direction and sense) to the
trihedral Oxyz (see § 391). It follows that D =±1 and
h = ± {rmnz — mz7i2), mi = ± {n2h — n^h), Wi = ± (^2^3 — ^3^2),
?2 =± (W3W1 — miWs), W2 =± (WsZi - W1Z3), W2 = ± (^3^11 — Zim3),
l3=± (miW2 — m2ni), W3 =± (nih — W2?i), m =± {hm2— hmi),
the upper or lower signs to be used according as the trihedrals are super-
posable or not.
391. A rectangular trihedral Oxijz is called right-handed if the rotation
that turns Oy through 90° into Oz appears counterclockwise as seen from
Ox ; otherwise it is called left-handed. In the present work right-handed
sets of axes have been used throughout.
Two right-handed as well as two left-handed rectangular trihedrals are
superposable ; a right-handed and a left-handed trihedral are not super-
posable. The difference is of the same kind as that between the gloves
of the right and left hand.
Two non-superposable rectangular trihedrals become superposable upon
reversing one (or all three) of the axes of either one.
XVI, § 393] QUADRIC SURFACES 355
392. The fact that the nine direction cosines are connected by six rela-
tions (§ 388) suggests that it must be possible to determine the position of
the new trihedral with respect to the old by only three angles. As such
we may take, in the case of superposable trihedrals, the angles 0, 0, \}/,
marked in Fig. 160, which are known as Eulefs angles.
The figure shows the intersections of the two trihedrals with a sphere
of radius 1 described about the origin as center. If OiVis the intersection
of the planes Oxy and Ox'y', Euler's angles are defined as
d = zOz', <t> = NOx', \p = xON.
The line ON is called the line of nodes^ or the nodal line.
Imagine the new trihedral Ox'y'z' initially coincident with the old
trihedral Oxyz, in direction and sense. Now turn the new trihedral
about Oz in the positive (counterclockwise) sense until Ox' coincides with
the assumed positive sense of the i^odal line ON; the amount of this
rotation gives the angle \^. Next turn the new trihedral about ON in the
positive sense until the plane Ox'y' assumes its final position ; this gives
the angle 6 as the angle between the planes Oxy and Ox'y\ or the angle
zOz' between their normals. Finally a rotation of the new trihedral
about the axis Oz\ which has reached its final position, in the positive
sense until Ox' assumes its final position, determines the angle 0.
393. The relations between the nine direction cosines and the three
angles of Euler are readily found from Fig. 150 by applying the fundamen-
tal formula of spherical trigonometry cos c = cos a cos 6 + sin a sin h cos 7
successively to the spherical triangles
xNx'^ xNy', xNz'^
yNx', yNy', yNz',
zNx'^ zNy'^ zNz'.
We find in this way :
li = cos xj/ cos — sin ^ sin (p cos 6,
mi = sin \{/ cos <f) -f cos xj/ sin cos ^,
wi = sin sin 0,
?2 = — cos i/' sin — sin \// cos cos 0, Z3 = sin ^ sin 0,
m2=— sin 1^ sin + cos ^ cos cos 0, mz=— cos \p sin 0^
ni = cos sin 0, m = cos 0.
APPENDIX
NOTE ON ABRIDGED NUMERICAL MULTIPLICATION
AND DIVISION
1. In multiplying two numbers it is convenient to write the
multiplier not below but to the right of the multiplicand in
the same line with it, and to begin the formation of the par-
tial products with the highest figure (and not with the lowest).
The most important part of the product is thus obtained first.
The partial products must then be moved out toward the right
(and not to the left). Thus :
35702
285616
24991 4
71
17
87025
404
8510
310696 6550
2. "Long" multiplications like the above rarely occur in
practice. Generally we have to multiply two numbers known
only approximately, to a certain number of significant figures.
Suppose we want to find the product of 3.5702 and 8.7025, five
significant figures only being known. It is then useless to
calculate the figures to the right of the vertical line in the
scheme above. To omit this useless part we proceed as fol-
lows. In multiplying by 8, place a dot over the last figure 2
of the multiplicand ; in multiplying by 7, place a dot over the
of the multiplicand, beginning the multiplication with this
figure (adding, however, the 1 which is to be carried from the
preceding product 7x2); then to indicate the multiplication
by simply place a dot over the 7 of the multiplicand; the
356
APPENDIX
357
multiplication by 2 has then to begin at the 5 of the multipli-
cand. Thus we obtain :
3.5702 1 8.7025
28 5616
2 4991
71
18
31.0696
The last figure so found is slightly uncertain, just as the
last figures of the^given numbers generally are.
3. In division it is most convenient to place th^^ivisor to
the right of the dividend. Thus
3.1416 =-8.90702
27.9823
25 1328
2 8495
2 8274
220
219
4
600
912
68800
62832
To cut off the superfluous part to the right of the vertical
line, subtract the first partial product as usual ; then cut off
the last figure from the divisor and divide by the remaining
portion ; go on in this way, cutting off a figure from the divisor
at every new division until the divisor is used up. Thus :
27.9823 |3J^ = 8.90701
25 1328
2 8495
2 8274
221
220
1
ANSWERS
[Answers which might in any way lessen the vahie of the Exercise are not
given.]
Pages 9-10. 5. 2| miles. 16. 173.9 ft.
Pages 13-14. 3. 22. 4. ^(pc + c« + ab).
7. K«^ + 2 6c - 2 ca - b'^) = K« - &)(« + b -2c).
Pages 17-18. 4. |rir2 sin (02 - 0i)-
5. I[r2r3 sin (03 - 02) + nn sin (0i - 03) + rir2sin (02- 0i)].
6. — ^'^^^cosi^(02 — 0i). 7. rcos0 = X + y cos w, rsin0 = ysin w.
•J'l + r2
Page 22. 17. They intersect at [\(xi-\-X2-{-x^ + Xi), ^(^1+^2+2/3+2/4)].
20. [i(a;i +X2 + X3), i(yi + 2/2 + 2/3)].
Page 35. 21. P = 1000(1 + r) ; P = 1000 + 60 n.
Page 38. 14. No.
Page 45. 1. (e) sin2j3; (/) a2a3 + «3«i + «ia2.
8. (&) (4, 3), (4, - 3), (- 4, 3), (- 4, - 3) ; (d) (3, - 2) ;
(e) (±i, ±3); (/) (t, i)-.
Pages 48-49. 1. (a) ; (6) ; (c) -113; (d) -5; (e) 1.
4. (a) (2, - 1, 3) ; (6) (83/41, - 81/41, - 35/41) ; (c) (- 5, 3, - 2) ;
id) (±3, i2, ±4); (e) (±1, ±1, ±1); (/) (1,0,-3).
Page 53. 1. (a) ; (6) - 180; (c) -27846; ((?) 7728; (e) 36;
(/) 550.
Page 57. 6. (27/2, -77/2).
Pages 59-60. 6.640/39. 9. (&1W2 - 62^1)2/2 mim2(wii - m2).
10. (3, i).
Pages 65-66. 2. (a) r sin = ± 5 ; (6) r cos = ± 4 ;
(c) rcos(0- f 7r)= ± 12.
3. = 0, rsin = 9, = ^ TT, r cos = 6. 14. 8464/85.
19. (- 5, - 10). 21. X = 1 (by inspection), 4x - 3y -\- 16 =0.
359
360 ANSWERS
Page 68. 4. K^ - ab = 0.
<
Page 69. 1. tan-i ^^^' ~ ^^ ; a = -b, h^=ab. '
a -i- b
4. [mi(62 — &)— W2(&i — 6)]^/2mim2(m2 — «ii).
6. r(2 cos - 3 sin 0) + 12 = 0.
10. 1 hr. 10 m. ; 176 miles from Detroit.
Page 75. 6. 560. 7. 120. 8. 65200. 9. 60 ; 24, 36.
10. 487635, 32509, 1653.
11. „C'i„, when n is even ; „C'|(^_j^ = «^^cn+i)' ^^^" " ^^ ^^^*
12. 66." 13. 120.
Pages 82-83. 2. aox^ + aix^ + a2X + as- 4. 8 abed.
6. (a) a; = 2, ?/=-!, 0=2, 10 = 3; (&) x = 1, y = 3, 5r = 2, lo = - 1.
7. (a) No; (b) Yes.
8. cos^ a + cos2 /3 + cos'^ y + 2 cos « cos /3 cos 7 pi-
pages 85-86. 2. (a) ABG+2FGH-AF^-BG-^-CH^]
(b) x^ + i/2 + 0-^ - 2(2/5 + 0X + xy) ; (c) - (x^ + i/^ + 0^) ; (e) 4.
7. («) l+a2 + 62 + c2. ^5) (a(^ + c/- &e)2 ; (c) (a(? + fte + c/)'^.
Pages 90-91. 6. x'^ + tf - 96x- 6iy ■{■ 2408 = ; 31.8 ft. or 66.3 ft.
8. a;2 + ?/2 - 16 X + 8 ?/ + 60 = 0. 9. A circle except for k =±l.
10. a;2 _|. y2 _,_ 4 L±A% 4- 4 = 0.
1 — k^
Page 92. 2. (a) 7-2-20 r sin 0+75=0 ;
(6) f^ -12 r cos (0 - i tt) 4- 18 = ; (c) r + 8 sin = 0.
Page 94. 8. ^2 - 6 a: + 28 = 0. 9. x'^ + 2 pwx + gm^ = 0.
Page 96. 3. (-6, -1), (29/106, 42/53).
7. 8a;-4?/-ll±15\/2 = 0. t^
Page 98. 3. (xi - h) {x - h) + {yi - k) (y - k) = r^.
7. i-r^A/C-rW/C), 8. (2,1).
Page 100. 6. {x - 79/38)2 + (y - 55/38)2 = (65/38)2.
8. ^2 + ?/2 + 4 X — 2 ?/ - 15 = 0.
Page 105. 1. (c) Polar lies at infinity.
Pages 108-109. 3. Let L, M be the intersections of the circle with
CPi, then ^2 ~ r2 = LPi - MPi.
ANSWERS 361
6. (c) 2x24-2y2_|.22x+6?/+15=0, 2x2+2?/2_i0x-10?/-25=0.
12. If the vertices of the square are (0, 0), (a, 0), (0, a), (a, a) and A;2
is the constant, the locus is 2 x2 + 2 ?/2 — 2 ax - 2 ai/ + 2 a2 - ^2 = ;
^•>a; |aV6.
13. If the vertices of the triangle are (a, 0), (—a, 0), (0, aV3) and
A;2 is the constant, the locus is 3 x2 + 3 y2 _ 2 VS a?/ + 3 a2 - 2 A:2 = 0.
Page 126. 8. (a) (3 + 4i)/25; (&) (3 + VrO/14 ;
(c) (-6 + 30/34; (c?) (1-6 0/37.
Page 130. 7. (^) ±K^^+v'20; {h.) v^2(cos80°+isin80°),
\/2(cos 200° + i sin 200=), ^2(cos 320° + i sin 320°).
Pages 135-136. 10. (a) 2 ?/ = 3 x2 + 5 x ;
(6) 12 ?/ = - 5 x2 + 29 X - 18.
11. 300 y = - x2 + 230 x ; 44.1 ft. above the ground ; 230 ft. from the
starting point.
20. (6) No parabola of the form y = ax^ + 6x + c is possible.
Page 138. 13. (2,3), (-1.8,3.6), (3.1, -2,8), (-3.3,-3.8).
Page 142. 6. East, East 33° 41' North, East 53° 8' North, East 18°
26' South.
10. 100/(9r+4).
Pages 145-146. 10. 0, 8° 8'. 11. 7° 29'.
15. When the side of the square is 3 in.
18. (a) 6 y = x8 + 6 2ic - 19 X ; (6) 7 y = 2 x^ - x2 - 29 x + 36.
Page 147. 1. (a) -1, 3.62, 1.38; (6) -1.45, -.403, .855 ;
(c) -1.94, .558, 1.38; (d) 2.79.
Page 154. 4. (d) -252xM; (d) ^0 a%^ - SO a^b^ ; (h) 27/a25.
Page 159. 3. (a) PiP2=Ps; (&) Pi^Ps=P2^ ; (c) pi^=27 p2^=7292-)s\
Page 162. 1. - 1.88, 1.53, .347.
Page 167. 1. (a) 4.06155 ; (b) ±2.08779; (c) 1.475773.
2. 2.0945514. 3. .34899.
4. (a) (1.88, 3), (- 1.53, 3), (- .347, 3) ;
(b) (.309, 1.10), (1.65, 1.55), (-1.96, .347) ; (c) (-2.106, -1.0266).
5. 3.39487 in. 6. 9.69579 ft. 7. - 2, 1 ± VS.
8. .22775, 3.1006. 9. 5.4418 ft.
10. (2, 3), (- 1.848, 3.584), (3.131, - 2.805), (- 3.383, - 3.779).
11. (2.21, .89). 12. .34729 a.
362 ANSWERS
Pages 173-174. 2. (a) (4, i tt), (4, | tt) ; (b) (a, ^ tt), (a, | tt) ;
(c)(4, 0); (d) (4aa'r), (4 a, fTr).
7. (a) 2/2 - 4 x + 4 = ; (6) 14 2/2 - 45 X + 52 ?/ + 60 = 0.
8. (6) a;2 - 10 a; - 3 2/ + 21 = ; (c) ^2 + 2 a; + y - 1 = 0.
9. The equation of a parabola contains an xy term when its axis is oblique
to a coordinate axis.
Pages 179-180. 1. (a) 18 a; - 30 ;
(5) 6 x5 - 30 X* + 48 x3 - 24 ic2 + 8 X - 8.
2. (a) y'=5/2y; (b) y' = 6/(5 - 2 y) ; (c) y' = 2/Sy.
5. (a)y'=-y/x; (b) y' = (6 - 2 xy) /x^ ;
(c) 2/'=-(^x + iry + G^)/(£rx + 5y + i?').-
Pages 186-188. 8. («) 2/=0 ; (6) 2a:+2i/-9=0, 2a;-|/-18=0;
(c) 2 X + 2 y - 9 = 0, 8 X + 16 y - 27 = 0, 24 X - 16 2/ - 153 = ;
(d) 8 X - 16 2/ - 27 = 0.
14. Directrix. 15. y'^ = a{x-3a). 22. 1^(1 + m2).
m2
29. a:2-80x-2400?/ = 0; 0, - |, - |, - i 0, f, 2.
30. x2 = 360(2/ -20).
Pages 194-195. 2. (3 7r-4)/6 7r. 3. § a2 (1±!^.
8. («) 64/3; (6) 625/12; (c) 1/12. 9. 123.84 ft^. 10. 1794i tons.
11. 199.4 ft2.
Page 197. To obtain the following solutions, take the origin at one
end of the beam and the axis Ox along the beam.
1. F- W, M= W(x-l). 2. F=io(^l-x), M:= ^w(l-x)x,
3. (a) Fi=-wx, Mi = -^wx^; F2=w{ll—x), M2=-iw{^P—lx+x^);
Fi = w{l-x), Mz=-\w{l-xy)
{b) Fi=-W, Mi=-Wx', F2 = 0', M.2=-\Wl]
Fs= W,M3=-W(il-x).
4. (a) Fi = lwl, Mi = lwlx', F2 = io(^l — x),
M2=-^ w?(a;2 -lx-\-\l^); i^s = - i wjZ, ^3 = z '^K^ - ^)'
Page 200. 9. 8^2 - 2 xy + 8 y2_ 63 = 0.
Page 204. 10. 3 x2 - ?/2 = 3 a^. 11. b. 14. 2 xy = 1.
Pages 211-212. 2. ^X4-^V=c2. 13. 54.5 ft., 42.2 ft. 18.62/^2.
x y
23. An ellipse or hyperbola according as one circle lies within or without
the other circle.
ANSWERS 363
Pages 221-222. 7. (a) A'a^ - B^b^ = C^ ;
(&) rt2cos2/3- &2sin2/3=i)2.
19. 62, 21. a2 + 62 . ^2 _ 62.
22. 4ab. 23. sin-i (a6/a'6')-
25. (a) a;2 + 2/2 = oj2 + 62 ; (&) ^2 + 2/2 = ^2 _ ^2.
Page 22?! 3. (a) (1, -1), (1±V2, -1), a;=li|V2;
(&) (i,0), (1,0), (-|,0),:r. = 0,a; = l.
4. 2 62/a. 8. (a) a2j,2 ^^ ft2j;(« _ a^) ; (&) b'^x^ = a^y{b - y).
10. Two straight lines.
Page 235. 2. (a) Vertices (5, 3), (8, 3); semi-axes 3/2, V2.
(6) Vertices (4, 8/3), (8, 8) ; semi-axes 10/3, 5\/3/3.
(c) vertices (17/6, 7/5), (1, 3) ; semi-axes \/65/5, \/l3/2.
3. 3a: + 2?/-2=0; (21/13, -37/26), 10/Vl3.
Page 237. 5. (acosd, — asind), x^ -{- y'^ — 2 a(xcose — ysine) = 0.
Pages 246-247. 2. (a) 3x-142/=0; (b) y = -S/l3, x=-14/13.
6. 2 x2 - xy - 15 2/2 -I- X + 19 2/ - 6 = 0,
2 x2 - iC2/ - 15 2/2 + X -^ 19 2/ - 28 = 0.
6. 6 x2 + icj/ - 2 2/2 - 9 x -t- 8 2/ - 46 = 0,
6 a;2 -H a;2/ - 2 y2 _ 9 a; + 8 2/ + 34 = 0.
11. (a) a:2/4 + y^ = 1 ; (6) x2/4 - 2/V2 = 1 ; (c) 3 x2 + ^2 _,_ ^ = ;
((?) a;2/16 +_2/V4 = 1 ; _(e) (3 + Vl7)x2 -}- (3 - Vl7)2/2 = 4 ;
(/) (2-hV2)x2-H(2-V2)y2 = l.
15. x^ + yi= a^.
19. Equilateral hyperbola.
Page 253. 2. («) Simple point ; (6) node ; (c) cusp ; (d) cusp.
4. (a) None ; (6) node at (6, 0) ; (c) isolated point at (a, 0) ;
(d) cusp at (a, 0).
Page 260. 4. r = a(sec <f> ± tan 0) or (x — a)y2 _^ 3.2(-a. ^_ q,) _ q.
10. x22/2 zz a2(a;2 4. ^2). 11. Cissoid (a - x)y^ = x^.
12. 2/(x2-l-2/2) = a(x2-2/2). 13. r = actn0.
14. (x2 -H 2/2)2 _ 4 «a;(x2 - 2/2).
Page 283. 6. ^ + ^^ etc.
V2(l + ZZ' -l-mm' + 7i«')
13. ^ (xi + X2 + X3), i (yi + ^2 + ys), i (^1 + 2^2 + Z3).
Page 287. 6. cos-i (7/3V29).
364
ANSWERS
Page 291. 2. ^V465.
3.
^269.
6. (3962, 47^ 43', 276° 16'), (320, - 2914, 2666), 2931.
7. I riViVl —[cos di cos 62 + sin Oi sin 62 cos (<pi — 02)]"^
02)+ cosfli cos ^2].
8. y/ri^ + rz^ — 2 rir2 [sin di sin ^2 cos (0i
10. - 1, 10, 7.
Page 296. 3. 39 a: - 10 ?/ + 7 - 89 = 0.
5. 97/28, - 97/49, - 97/9. 7. Sx - 4:y + 2 z- 6 = 0.
Page 300. 5. 4ic + 8?/ + 2; = 81, 4x-|-8a; + ^ = 90.
Page 303. 2. (a) 56/3; (ft) 0; (c) 19/3.
Page 306. 12. Sx-2y = l. 13. 6a; + 11 ?/ + 90 = 58.
16. 70° 31'. 17. cos-i(2/i-^ + 3rt2)/(4/i2_|.3«2).
Pages 314-316. 3. 69° 29'. 19. (a) V637l9; (ft) V194/33.
21. X - 2 y + ^ + 8 = 0.
X2 — Xi 2/2 — 2/1 ;22 - i^l
24. ai fti ci =0.
a2 ft2 C2
Page 320. 11. ( - 3, - 3, 2), (9, 9, - 6).
Pages325-326. 4. (1, 0, - 3), (- 9/11, 20/11, 27/11).
7. a:2 - 3 y^ - S z^ = 0. 13. 25(^2 + y^ + z^) = 16^, 25 = 64.
Pages 329-331. 4. (4,-5,-3). 5. (4,6,2).
6. ^x + 2y - z = 25,2x~3y -\-z + 25 = 0.
20. 9ic2 + 4?/2 + 1.3 2;2 + 2 X2/ - 273 = 0.
21. (x - ZA:)2 + (^ - mky^ + (s - wA;)2 = r2.
22. ll{x-\-h)+m{y-\-j) + n{zi-k)Y-[{x+hy-\-{y-hj)^+(z + k)^-r'2]=0.
Page 336. 3. Va^ - c2 x ± Vft2 - c2 = 0.
6. (a;2 + ^2 + ;22 _ «2 _ ft2)2 _ 4 52(q;2 _ ^2) ^ Q.
8. (a) 16a2(a;2 4-;32) = y4. (&) 16 a2[(x + a)2 + 02] = (4a2 - y2)2.
9. y^{x^ + z^)=a^. ^
INDEX
{The numbers refer to the pages.)
Abscissa, 1, 4.
Absolute value, 124.
Acnode, 252.
Adiabatic expansion, 276.
Algebraic curves, 249-253.
Amplitude, 16, 124.
Angle between line and plane, 312 ;
between two lines, 58, 284, 311;
between two planes, 299.
Anomaly, 16.
Area of ellipse, 221 ; of parabolic
segment, 191-195; of triangle, 11,
12, 56, 288 ; under any curve, 193.
Argument, 124.
Associative law, 110.
Asymptotes, 203.
Axes of coordinates, 4, 277 ; of ellipse,
198 ; of hyperbola, 202.
Axis, 18 ; of parabola, 132, 170 ; of
pencil, 303 ; of symmetry, 137.
Azimuth, 16.
Bending moment, 196-197.
Binomial coefficients, 152-154 ; the-
orem, 152-154.
Bisecting planes, 299.
Bisectors of angles of two lines, 64.
Cardioid, 255.
Cartesian coordinates, 16.
Cartesian equation of conic, 225 ; of
ellipse, 199 ; of hyperbola, 202 ; of
parabola, 171.
Cartesius, 17.
Cassinian ovals, 256, 259.
Catenary, 188.
Center of ellipse, 198, 215; of hyper-
bola, 202, 215 ; of inversion, 101 ;
of pencil, 67; of sheaf, 304; of
symmetry, 137.
Centroid, 22.
Chord of contact, 103.
Circle, 87-109 ; in space, 321.
Circular cone, 341.
Cissoid, 255.
Classification of conies, 225 ; of
quadric surfaces, 342-345.
Clockwise, 11.
Cofactors, 52, 80.
Colatitude, 290.
Column, 41, 47.
Combinations, 73-75.
Common chord, 107 ; logarithms, 264.
Commutative law, 110.
Completing the square, 88, 133.
Complex numbers, 100, 115, 117-130.
Component, 19, 280.
Conchoid, 254.
Cone, 341, 351 ; of revolution, 342.
Conic sections, 223-231, 232.
Conies as sections of a cone, 228-231.
Conjugate axes, 327 ; axis, 203 ; com-
plex numbers, 122 ; diameters, 215-
219 ; elements of determinant,
83; lines, 327.
Continuity, 155-156.
Contour lines, 351.
Coordinate axes, 4, 277 ; planes, 277 ;
trihedral, 277.
Coordinates, 1, 5, 277 ; polar, 16, 290.
Cosine curve, 261.
Counterclockwise, 11.
Cross-sections, 333, 337, 339, 350.
Crunode, 252.
Cubic curves, 248 ; equation, 146-
147; function, 143-147.
Curve in space, 293.
Cusp, 252.
Cycloid, 257.
Cylinders, 351.
365
366
INDEX
De Moivre's tlieorem, 126.
Derivative, 139-141, 143, 149-152,
177-179; of ax^, 139; of cubic
function, 143 ; of function of a func-
tion, 178 ; of implicit function, 177-
179 ; of polynomial, 149-151 ; of
product, 178 ; of quadratic func-
tion, 140; of a:", 151.
Descartes, 17.
Determinant, 11, 13, 39; of n
equations, 81 ; of order n, 77 ; of
second order, 41 ; of three equa-
tions, 48 ; of third order, 47 ; of
two equations, 41.
Diameter, 333; of ellipse, 215; of
hyperbola, 218; of parabola, 184-
185.
Direction cosines, 282, 307.
Director circle, 222 ; sphere, 349.
Directrices of conies, 223, 226.
Directrix of parabola, 169.
Discriminant of equation of second
degree, 240-241 ; of quadratic
equation, 92.
Distance between two points, 7, 17,
278 ; of point from line, 63, 313 ;
of point from origin, 6, 278 ; of
point from plane, 298 ; of two
lines, 313-314.
Distributive law, 110.
Division, abridged, 357.
Division ratio, 3, 8, 281.
Double point, 251.
Eccentric angle, 220.
Eccentricity, 208, 223.
Elements of determinant, 47; of
permutations and combinations, 70.
Elimination, 43, 54, 82.
EUipse, 198-222, 223, 229, 242-244.
Ellipsoid, 332-334 ; of revolution, 334.
Elliptic cone, 341 ; paraboloid, 340.
Empirical equations, 266-276.
Epicycloid, 258.
Equation of first degree, see Linear
equation ; of line, 26, 32 ; of plane,
293-297 ; of second degree, 88.
Equations of line, 308.
Equator, 334.
Equatorial plane, 290.
Equilateral hyperbola, 203.
Euler's angles, 355.
Expansion by minors, 51, 80.
Explicit and implicit functions, 177.
Exponential curve, 263.
Factor of proportionality, 25.
Factorial, 71.
Falling body, 15, 31, 69, 134.
Family of circles, 107 ; of spheres,
329.
Foci of conic, 226; of ellipse, 198,
223 ; of hyperbola, 201, 223.
Focus of parabola, 169.
Four-cusped hypocycloid, 259.
Function, 29 ; of two variables, 351.
Fundamental laws of algebra, 110.
Gas-meter, 27, 269.
Gas pressure, 272, 276.
General equation of second degree,
88, 233-247, 317, 342.
Geometric representation of complex
numbers, 117.
Higher plane curves, 248-276.
Homogeneous function of second
degree, 241 ; linear equations, 43,
54.
Hooke's law, 15, 25, 30, 38, 267,
269.
Horner's process, 166.
Hyperbola, 201-222, 223, 230, 242-
244.
Hyperbolic logarithms, 264 ; para-
boloid, 340 ; spiral, 259. ^
Hyperboloid, of one sheet,' 337-338 ;
of revolution of one sheet, 338 ; of
revolution of two sheets, 339 ; of
two sheets, 338-339.
Hypocycloid, 259.
Imaginary axis, 117; ellipsoid, 339;
numbers, 115; roots, 127, 160;
unit, 115 ; values in geometry, 116.
Implicit functions, 177.
Inclined plane, 271.
Induction, mathematical, 71.
Inflection, 144.
Intercept, 26, 34 ; form, 33, 295.
Interpolation, 161.
Intersecting lines, 307.
INDEX
367
Intersection of line and circle, 95 ;
of line and ellipse, 213 ; of line and
parabola, 181 ; of line and sphere,
323 ; of two lines, 39, 43.
Inverse of a circle, 101 ; operations,
111; trigonometric curves, 261-
262.
Inverses of involution, 112.
Inversion, 100, 324.
Inversions in permutations, 75.
Inversor, 109.
Irrational numbers, 113.
Isolated point, 252.
Latitude, 290.
Latus rectum of parabola, 170 ; of
conic, 224.
Laws of algebra, 110; of exponents,
112.
Leading elements, 83.
Left-handed trihedral, 354.
Lemniscate, 257, 260.
Level lines, 351.
LimaQon, 254.
Limiting cases of conies, 230.
Line, 24, 307 ; and plane perpendic-
ular at given point, 312 ; of nodes,
355 ; parallel to an axis, 23 ; through
one point, 36, 308 ; through origin,
24 ; through two points, 36, 56,
308.
Linear equation, 32, 293.
Linear equations, n, 81 ; three, 46,
48, 302 ; two, 39-42, 293, 302.
Linear function, 29, 131.
Lituus, 259.
Logarithm, 263-265.
Logarithmic paper, 274; plotting,
272-276.
Longitude, 290.
Major axis, 199.
Mathematical induction, 71.
Maximum, 141, 143.
Measurement, 114.
Mechanical construction of ellipse,
198; of hyperbola, 201; of parab-
ola, 171.
Melting point of alloy, 175, 269.
Meridian plane, 290; section, 335.
Midpoint of segment, 9.
Minimum, 141, 143.
Minor axis, 199.
Minors of determinant, 51, 80.
Modulus of complex number, 124 ;
of logarithmic system, 265.
Moment of a force, 288.
Multiple points, 253.
Multiplication, abridged, 356.
Multiplication of determinants, 84.
Napierian logarithms, 264.
Natural logarithms, 264.
Negative roots, 166.
Newton's method of approximation,
162.
Nodal line, 355.
Node, 252.
Non-linear equations representing
lines, 68.
Normal form, 61, 296.
Normal to ellipse, 208 ; to parabola,
181, 182 ; to any surface, 349.
Numerical equations, 158-168.
Oblate, 334.
Oblique axes, 6, 7, 38, 278.
Octant, 277.
Ordinary point, 251.
Ordinate, 5.
Origin, 1, 4, 277.
Orthogonal substitution, 352 ; trans-
formation, 352.
Parabola, 131-142, 169-197, 229,
244-245; Cartesian equation, 171 ;
polar equation, 169-170 ; referred
to diameter and tangent, 190.
Paraboloid, elliptic, 340 ; hyper-
bolic, 340 ; of revolution, 341.
Parallel, 335 ; circle, 335.
Parallelism, 28, 33, 59, 285.
Parallelogram law, 19, 120.
Parameter, 107, 109 ; equations of
circle, 109; of ellipse, 220; of
hyperbola, 220 ; of parabola, 189.
Pascal's triangle, 154.
Peaucellier's cell, 109.
Pencil of circles, 107 ; of lines, 67 ;
of parallels, 67 ; of planes, 303 ; of
spheres, 329.
Pendulum, 134,
368
INDEX
Permutations, 70-73.
Perpendicularity, 28, 33, 59, 285.
Phase, 124.
Plane, 292-306 ; through three points,
295.
Plotting by points, 131.
Points of inflection, 144.
Polar, 102, 104, 326 ; angle, 16 ; axis,
16 ; coordinates, 16, 290 ; equa-
tion of circle, 91 ; of conic, 224-225;
of line, 60 ; of parabola, 169-170 ;
representation of complex num-
bers, 124.
Pole, 16.
Pole and polar, 102, 104, 326.
Poles, 334.
Polynomial, 148-157 ; curve, 155-157.
Power of a point, 106, 328.
Principal diagonal, 47.
Projectile, 135, 142.
Projecting cylinders, 321 ; planes
of a line, 309-311.
Projection, 18-21, 280-281, 284.
Prolate, 334.
Proportional quantities, 24.
Pulleys, 27, 31, 38, 268.
Pythagorean relation, 282.
Quadrant, 5.
Quadratic equation, 92 ; function,
131-142.
Quadric surfaces, 332-^50, 342.
Radical axis, 106, 328, 329 ; center,
107, 328, 329 ; plane, 328.
Radius vector, 16, 282, 290.
Rate of change, 29, 149 ; of interest,
29, 35.
Rational numbers. 111.
Real axis, 117; numbers, 113; roots,
160-167.
Reciprocal polars, 327.
Rectangular coordinates, 6 ; hyper-
bola, 203.
Reduction to normal form, 62, 297.
Regula falsi, 161.
Related quantities, 14.
Remainder theorem, 163.
Removal of term in xy, 238.
Resultant, 19, 280.
Right-handed trihedral, 354.
Rotation of axes, 235-236; of co-
ordinate trihedral, 352-355.
Row, 41, 47.
Rule of false position, 161.
Ruled surfaces, 347-349.
Rulings on hyperboloid of one sheet,
348 ; on hyperbolic paraboloid, 349.
Second derivative, 144.
Secondary diagonal, 47.
Sheaf of planes, 304.
Shearing force, 196-197.
Shortest distance of two lines, 313-
314.
Simple point, 251.
Simpson's rule, 193.
Simultaneous linear equations, 39-
48, 81-83, 302.
Simultaneous linear and quadratic
equations, 94.
Sine curve, 261.
Skew symmetric determinant, 84.
Slope, 24 ; of ellipse, 207 ; of hyper-
bola, 210; of parabola, 139-140,
176 ; of secant of parabola, 138.
Slope form of equation of line, 26.
Sphere, 317-331 ; through four points,
319.
Spherical coordinates, 290.
Spheroid, 334.
Spinode, 252.
Spiral of Archimedes, 259.
Square root of complex number, 129.
Statistics, 14.
Straight line, 23.
Strophoid, 260.
Subnormal to parabola, 181.
Substitutions, 270.
Subtangent to parabola, 180.
Sum of two determinants, 52, 78.
Superposable trihedrals, 354.
Surface, 292 ; of revolution, 335-336.
Suspension bridge, 188.
Symmetric determinant, 84.
Symmetry, 136-138, 215.
Synthetic division, 164.
Tangent to algebraic curve at origin,
250-253; to circle, 97; to ellipse,
206, 213; to hyperbola, 210; to
parabola, 139, 180, 182.
INDEX
369
Tangent cone to sphere, 324.
Tangent curve, 261.
Tangent plane to ellipsoid, 346 ; to
hyperboloids, 347 ; to paraboloids,
347 ; to quadric surfaces, 347 ; to
sphere, 322.
Taylor's theorem, 168.
Temperature, 15, 31, 270.
Tetrahedron volume, 301.
Thermometer, 2, 31, 35.
Transcendental curves, 262.
Transformation from cartesian to
polar coordinates, 16, 290-291;
to center, 226, 240; to parallel
axes, 12, 239.
Translation of axes, 12, 233-235 ; of
coordinate trihedral, 287.
I Transposition, 50, 78.
Transverse axis, 203.
Trochoid, 258.
Uniform motion, 30, 69.
Units, 5.
Vector, 18, 119, 280.
Vectorial angle, 16.
Velocity, 30, 31.
Versiera, 256.
Vertex of parabola, 132, 170.
Vertices of ellipse, 198 ; of hyper-
bola, 202.
Volume of tetrahedron, 301.
Water gauge, 2.
Whispering galleries, 212.
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